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B O U N D A R YVALUE PROBLEMS

FIFTH EDITION

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B O U N D A R YVALUE PROBLEMSANDPARTIAL DIFFERENTIAL EQUATIONS

DAVID L. POWERSC l a r k s o n U n i v e r s i t y

FIFTH EDITION

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Contents

Preface ix

CHAPTER 0 Ordinary Differential Equations 10.1 Homogeneous Linear Equations 10.2 Nonhomogeneous Linear Equations 140.3 Boundary Value Problems 260.4 Singular Boundary Value Problems 380.5 Green’s Functions 43

Chapter Review 51Miscellaneous Exercises 51

CHAPTER 1 Fourier Series and Integrals 591.1 Periodic Functions and Fourier Series 591.2 Arbitrary Period and Half-Range Expansions 641.3 Convergence of Fourier Series 731.4 Uniform Convergence 791.5 Operations on Fourier Series 851.6 Mean Error and Convergence in Mean 901.7 Proof of Convergence 951.8 Numerical Determination of Fourier Coefficients 1001.9 Fourier Integral 1061.10 Complex Methods 1131.11 Applications of Fourier Series and Integrals 1171.12 Comments and References 124


v

vi Contents

CHAPTER 2 The Heat Equation 1352.1 Derivation and Boundary Conditions 1352.2 Steady-State Temperatures 1432.3 Example: Fixed End Temperatures 1492.4 Example: Insulated Bar 1572.5 Example: Different Boundary Conditions 1632.6 Example: Convection 1702.7 Sturm–Liouville Problems 1752.8 Expansion in Series of Eigenfunctions 1812.9 Generalities on the Heat Conduction Problem 1842.10 Semi-Infinite Rod 1882.11 Infinite Rod 1932.12 The Error Function 1992.13 Comments and References 204


CHAPTER 3 The Wave Equation 2153.1 The Vibrating String 2153.2 Solution of the Vibrating String Problem 2183.3 d’Alembert’s Solution 2273.4 One-Dimensional Wave Equation: Generalities 2333.5 Estimation of Eigenvalues 2363.6 Wave Equation in Unbounded Regions 2393.7 Comments and References 246


CHAPTER 4 The Potential Equation 2554.1 Potential Equation 2554.2 Potential in a Rectangle 2594.3 Further Examples for a Rectangle 2644.4 Potential in Unbounded Regions 2704.5 Potential in a Disk 2754.6 Classification and Limitations 2804.7 Comments and References 283


CHAPTER 5 Higher Dimensions and Other Coordinates 2955.1 Two-Dimensional Wave Equation: Derivation 2955.2 Three-Dimensional Heat Equation 2985.3 Two-Dimensional Heat Equation: Solution 303

Contents vii

5.4 Problems in Polar Coordinates 3085.5 Bessel’s Equation 3115.6 Temperature in a Cylinder 3165.7 Vibrations of a Circular Membrane 3215.8 Some Applications of Bessel Functions 3295.9 Spherical Coordinates; Legendre Polynomials 3355.10 Some Applications of Legendre Polynomials 3455.11 Comments and References 353


CHAPTER 6 Laplace Transform 3636.1 Definition and Elementary Properties 3636.2 Partial Fractions and Convolutions 3696.3 Partial Differential Equations 3766.4 More Difficult Examples 3836.5 Comments and References 389

Miscellaneous Exercises 389

CHAPTER 7 Numerical Methods 3977.1 Boundary Value Problems 3977.2 Heat Problems 4037.3 Wave Equation 4087.4 Potential Equation 4147.5 Two-Dimensional Problems 4207.6 Comments and References 428


Bibliography 433

Appendix: Mathematical References 435

Answers to Odd-Numbered Exercises 441

Index 495

Preface

This text is designed for a one-semester or two-quarter course in partial dif-ferential equations given to third- and fourth-year students of engineering andscience. It can also be used as the basis for an introductory course for graduatestudents. Mathematical prerequisites have been kept to a minimum — calculusand differential equations. Vector calculus is used for only one derivation, andnecessary linear algebra is limited to determinants of order two. A reader needsenough background in physics to follow the derivations of the heat and waveequations.

The principal objective of the book is solving boundary value problemsinvolving partial differential equations. Separation of variables receives thegreatest attention because it is widely used in applications and because it pro-vides a uniform method for solving important cases of the heat, wave, andpotential equations. One technique is not enough, of course. D’Alembert’s so-lution of the wave equation is developed in parallel with the series solution,and the distributed-source solution is constructed for the heat equation. Inaddition, there are chapters on Laplace transform techniques and on numeri-cal methods.

The second objective is to tie together the mathematics developed and thestudent’s physical intuition. This is accomplished by deriving the mathemati-cal model in a number of cases, by using physical reasoning in the mathemat-ical development, by interpreting mathematical results in physical terms, andby studying the heat, wave, and potential equations separately.

In the service of both objectives, there are many fully worked examples andnow about 900 exercises, including miscellaneous exercises at the end of eachchapter. The level of difficulty ranges from drill and verification of detailsto development of new material. Answers to odd-numbered exercises are in

ix

x Preface

the back of the book. An Instructor’s Manual is available both online and inprint (ISBN: 0-12-369435-3), with the answers to the even-numbered prob-lems. A Student Solutions Manual is available both online and in print (ISBN:0-12-088586-7), that contains detailed solutions of odd-numbered problems.

There are many ways of choosing and arranging topics from the book toprovide an interesting and meaningful course. The following sections formthe core, requiring at least 14 hours of lecture: Sections 1.1–1.3, 2.1–2.5, 3.1–3.3, 4.1–4.3, and 4.5. These cover the basics of Fourier series and the solutionsof heat, wave, and potential equations in finite regions. My choice for the nextmost important block of material is the Fourier integral and the solution ofproblems on unbounded regions: Sections 1.9, 2.10–2.12, 3.6, and 4.4. Theserequire at least six more lectures.

The tastes of the instructor and the needs of the audience will govern thechoice of further material. A rather theoretical flavor results from including:Sections 1.4–1.7 on convergence of Fourier series; Sections 2.7–2.9 on Sturm–Liouville problems, and the sequel, Section 3.4; and the more difficult parts ofChapter 5, Sections 5.5–5.10 on Bessel functions and Legendre polynomials.On the other hand, inclusion of numerical methods in Sections 1.8 and 3.5and Chapter 7 gives a very applied flavor.

Chapter 0 reviews solution techniques and theory of ordinary differentialequations and boundary value problems. Equilibrium forms of the heat andwave equations are derived also. This material belongs in an elementary differ-ential equations course and is strictly optional. However, many students haveeither forgotten it or never seen it.

For this fifth edition, I have revised in response to students’ changing needsand abilities. Many sections have been rewritten to improve clarity, provideextra detail, and make solution processes more explicit. In the optional Chap-ter 0, free and forced vibrations are major examples for solution of differentialequations with constant coefficients. In Chapter 1, I have returned to derivingthe Fourier integral as a “limit” of Fourier series. New exercises are includedfor applications of Fourier series and integrals. Solving potential problems on arectangle seems to cause more difficulty than expected. A new section 4.3 givesmore guidance and examples as well as some information about the Poissonequation. New exercises have been added and old ones revised throughout.In particular I have included exercises based on engineering research publica-tions. These provide genuine problems with real data.

A new feature of this edition is a CD with auxiliary materials: animationsof convergence of Fourier series; animations of solutions of the heat and waveequations as well as ordinary initial value problems; color graphics of solu-tions of potential problems; additional exercises in a workbook style; reviewquestions for each chapter; text material on using a spreadsheet for numericalmethods. All files are readable with just a browser and Adobe Reader, availablewithout cost.

Preface xi

I wish to acknowledge the skillful work of Cindy Smith, who was the LaTeXcompositor and corrected many of my mistakes, the help of Academic Presseditors and consultants, and the guidance of reviewers for this edition:

Darryl Yong, Harvey Mudd CollegeKen Luther, Valparaiso UniversityAlexander Kirillov, SUNY at Stony BrookJames V. Herod, Georgia Tech UniversityHilary Davies, University of Alaska AnchorageCatherine Crawford, Elmhurst CollegeAhmed Mohammed, Ball State University

I also wish to acknowledge the guidance of reviewers for the previous edi-tion:

Linda Allen, Texas Tech UniversityIlya Bakelman, Texas A&M UniversityHerman Gollwitzer, Drexel UniversityJames Herod, Georgia Institute of TechnologyRobert Hunt, Humboldt State UniversityMohammad Khavanin, University of North DakotaJeff Morgan, Texas A&M UniversityJim Mueller, California Polytechnic State UniversityRon Perline, Drexel UniversityWilliam Royalty, University of IdahoLawrence Schovanec, Texas Tech UniversityAl Shenk, University of California at San DiegoMichael Smiley, Iowa State UniversityMonty Strauss, Texas Tech UniversityKathie Yerion, Gonzaga University

David L. Powers

Ordinary DifferentialEquations C H A P T E R

0

0.1 Homogeneous Linear Equations

The subject of most of this book is partial differential equations: their physicalmeaning, problems in which they appear, and their solutions. Our principalsolution technique will involve separating a partial differential equation intoordinary differential equations. Therefore, we begin by reviewing some factsabout ordinary differential equations and their solutions.

We are interested mainly in linear differential equations of first and secondorders, as shown here:

du

dt= k(t)u + f (t), (1)

d2u

dt2+ k(t)

du

dt+ p(t)u = f (t). (2)

In either equation, if f (t) is 0, the equation is homogeneous. (Another test: Ifthe constant function u(t) ≡ 0 is a solution, the equation is homogeneous.) Inthe rest of this section, we review homogeneous linear equations.

A. First-Order EquationsThe most general first-order linear homogeneous equation has the form

du

dt= k(t)u. (3)

1

2 Chapter 0 Ordinary Differential Equations

This equation can be solved by isolating u on one side and then integrating:

1

u

du

dt= k(t),

ln |u| =∫

k(t)dt + C,

u(t) = ±eC e∫

k(t)dt = ce∫

k(t)dt . (4)

It is easy to check directly that the last expression is a solution of the differentialequation for any value of c. That is, c is an arbitrary constant and can be usedto satisfy an initial condition if one has been specified.

Example.Solve the homogeneous differential equation

du

dt= −tu.

The procedure outlined here gives the general solution

u(t) = ce−t2/2

for any c. If an initial condition such as u(0) = 5 is specified, then c must bechosen to satisfy it (c = 5). �

The most common case of this differential equation has k(t) = k constant.The differential equation and its general solution are

du

dt= ku, u(t) = cekt . (5)

If k is negative, then u(t) approaches 0 as t increases. If k is positive, then u(t)increases rapidly in magnitude with t. This kind of exponential growth oftensignals disaster in physical situations, as it cannot be sustained indefinitely.

B. Second-Order EquationsIt is not possible to give a solution method for the general second-order linearhomogeneous equation,

d2u

dt2+ k(t)

du

dt+ p(t)u = 0. (6)

Nevertheless, we can solve some important cases that we detail in what follows.The most important point in the general theory is the following.

Chapter 0 Ordinary Differential Equations 3

Principle of Superposition. If u1(t) and u2(t) are solutions of the same linearhomogeneous equation (6), then so is any linear combination of them: u(t) =c1u1(t) + c2u2(t). �

This theorem, which is very easy to prove, merits the name of principle be-cause it applies, with only superficial changes, to many other kinds of linear,homogeneous equations. Later, we will be using the same principle on partialdifferential equations. To be able to satisfy an unrestricted initial condition, weneed two linearly independent solutions of a second-order equation. Two so-lutions are linearly independent on an interval if the only linear combination ofthem (with constant coefficients) that is identically 0 is the combination with 0for its coefficients. There is an alternative test: Two solutions of the same linearhomogeneous equation (6) are independent on an interval if and only if theirWronskian

W(u1,u2) =∣∣∣∣∣u1(t) u2(t)

u′1(t) u′

2(t)

∣∣∣∣∣ (7)

is nonzero on that interval.If we have two independent solutions u1(t), u2(t) of a linear second-order

homogeneous equation, then the linear combination u(t) = c1u1(t) + c2u2(t)is a general solution of the equation: Given any initial conditions, c1 and c2 canbe chosen so that u(t) satisfies them.

1. Constant coefficientsThe most important type of second-order linear differential equation that canbe solved in closed form is the one with constant coefficients,

d2u

dt2+ k

du

dt+ pu = 0 (k,p are constants). (8)

There is always at least one solution of the form u(t) = emt for an appropriateconstant m. To find m, substitute the proposed solution into the differentialequation, obtaining

m2emt + kmemt + pemt = 0,

or

m2 + km + p = 0 (9)

(since emt is never 0). This is called the characteristic equation of the differ-ential equation (8). There are three cases for the roots of the characteristicequation (9), which determine the nature of the general solution of Eq. (8).These are summarized in Table 1.

This method of assuming an exponential form for the solution works forlinear homogeneous equations of any order with constant coefficients. In all


Roots of Characteristic General Solution of DifferentialEquation Equation

Real, distinct: m1 �= m2 u(t) = c1em1t + c2em2t

Real, double: m1 = m2 u(t) = c1em1t + c2tem1t

Conjugate complex: u(t) = c1eαt cos(βt) + c2eαt sin(βt)m1 = α + iβ,m2 = α − iβ

Table 1 Solutions of d2udt2 + k du

dt + pu = 0

cases, a pair of complex conjugate roots m = α ± iβ leads to a pair of complexsolutions

eαteiβt, eαte−iβt (10)

that can be traded for the pair of real solutions

eαt cos(βt), eαt sin(βt). (11)

We include two important examples. First, consider the differential equation

d2u

dt2+ λ2u = 0, (12)

where λ is constant. The characteristic equation is m2 + λ2 = 0, with rootsm = ±iλ. The third case of Table 1 applies if λ �= 0; the general solution of thedifferential equation is

u(t) = c1 cos(λt) + c2 sin(λt). (13)

Second, consider the similar differential equation

d2u

dt2− λ2u = 0. (14)

The characteristic equation now is m2 − λ2 = 0, with roots m = ±λ. If λ �= 0,the first case of Table 1 applies, and the general solution is

u(t) = c1eλt + c2e−λt . (15)

It is sometimes helpful to write the solution in another form. The hyperbolicsine and cosine are defined by

sinh(A) = 1

2

(eA − e−A

), cosh(A) = 1

2

(eA + e−A

). (16)

Thus, sinh(λt) and cosh(λt) are linear combinations of eλt and e−λt . By thePrinciple of Superposition, they too are solutions of Eq. (14). The Wronskian


Figure 1 Mass–spring–damper system.

test shows them to be independent. Therefore, we may equally well write

u(t) = c′1 cosh(λt) + c′

2 sinh(λt)

as the general solution of Eq. (14), where c′1 and c′

2 are arbitrary constants.

Example: Mass–Spring–Damper System.The displacement of a mass in a mass–spring–damper system (Fig. 1) is de-scribed by the initial value problem

d2u

dt2+ b

du

dt+ ω2u = 0,

u(0) = u0du

dt(0) = v0.

The equation is derived from Newton’s second law. Coefficients b and ω2

are proportional to characteristic constants of the damper and the spring, re-spectively. The characteristic equation of the differential equation is

m2 + bm + ω2 = 0,

with roots

−b ± √b2 − 4ω2

2= −b

2±

√(b

2

)2

− ω2.

The nature of the solution, and therefore the motion of the mass, is determinedby the relation between b/2 and ω.

b = 0: undamped. The roots are ±iω and the general solution of the differ-ential equation is

u(t) = c1 cos(ωt) + c2 sin(ωt).

The mass oscillates forever.


0 < b/2 < ω: underdamped. The roots are complex conjugates α± iβ withα = −b/2, β = √

ω2 − (b/2)2. The general solution of the differential equa-tion is

u(t) = e−bt/2(c1 cos(βt) + c2 sin(βt)

).

The mass oscillates, but approaches equilibrium as t increases.

b/2 = ω: critically damped. The roots are both equal to b/2. The generalsolution of the differential equation is

u(t) = e−bt/2(c1 + c2t).

The mass approaches equilibrium as t increases and may pass through equi-librium (u(t) may change sign) at most once.

b/2 > ω: overdamped. Both roots of the characteristic equation are real,say, m1 and m2. The general solution of the differential equation is

u(t) = c1em1t + c2em2t .

The mass approaches equilibrium as t increases, and u(t) may change sign atmost once. These cases are illustrated on the CD. �

2. Cauchy–Euler equationOne of the few equations with variable coefficients that can be solved in com-plete generality is the Cauchy–Euler equation:

t2 d2u

dt2+ kt

du

dt+ pu = 0. (17)

The distinguishing feature of this equation is that the coefficient of the nthderivative is the nth power of t, multiplied by a constant. The style of solutionfor this equation is quite similar to the preceding: Assume that a solution hasthe form u(t) = tm, and then find m. Substituting u in this form into Eq. (17)leads to

t2m(m − 1)tm−2 + ktmtm−1 + ptm = 0, or

m(m − 1) + km + p = 0 (k,p are constants). (18)

This is the characteristic equation for Eq. (17), and the nature of its roots de-termines the solution, as summarized in Table 2.

One important example of the Cauchy–Euler equation is

t2 d2u

dt2+ t

du

dt− λ2u = 0, (19)


Roots of Characteristic General Solution of DifferentialEquation Equation

Real, distinct roots: m1 �= m2 u(t) = c1tm1 + c2tm2

Real, double root: m1 = m2 u(t) = c1tm1 + c2(ln t)tm1

Conjugate complex roots: u(t) = c1tα cos(β ln t) + c2tα sin(β ln t)m1 = α + iβ,m2 = α − iβ

Table 2 Solutions of t2 d2udt2 + kt du

dt + pu = 0

where λ > 0. The characteristic equation is m(m−1)+m−λ2 = m2 −λ2 = 0.The roots are m = ±λ, so the first case of Table 2 applies, and

u(t) = c1tλ + c2t−λ (20)

is the general solution of Eq. (19).For the general linear equation

d2u

dt2+ k(t)

du

dt+ p(t)u = 0,

any point where k(t) or p(t) fails to be continuous is a singular point of thedifferential equation. At such a point, solutions may break down in variousways. However, if t0 is a singular point where both of the functions

(t − t0)k(t) and (t − t0)2p(t) (21)

have Taylor series expansions, then t0 is called a regular singular point. TheCauchy–Euler equation is an example of an important differential equationhaving a regular singular point (at t0 = 0). The behavior of its solution nearthat point provides a model for more general equations.

3. Other equationsOther second-order equations may be solved by power series, by change ofvariable to a kind already solved, or by sheer luck. For example, the equation

t4 d2u

dt2+ λ2u = 0, (22)

which occurs in the theory of beams, can be solved by the change of variables

t = 1

z, u(t) = 1

zv(z).


Here are the details. The second derivative of u has to be replaced by its ex-pression in terms of v, using the chain rule. Start by finding

du

dt= d

dz

(v

z

)· dz

dt.

Since t = 1/z, also z = 1/t, and dz/dt = −1/t2 = −z2. Thus

du

dt= −z2

(zv′ − v

z2

)= −zv′ + v.

Similarly we find the second derivative

d2u

dt2= d

dz

(du

dt

)dz

dt= d

dz(−zv′ + v)

(−z2)

= −z2(−zv′′ − v′ + v′) = z3v′′.

Finally, replace both terms of the differential equation:(1

z

)4

z3v′′ + λ2 v

z= 0,

or

v′′ + λ2v = 0.

This equation is easily solved, and the solution of the original is then found byreversing the change of variables:

u(t) = t(c1 cos(λ/t) + c2 sin(λ/t)

). (23)

C. Second Independent SolutionAlthough it is not generally possible to solve a second-order linear homoge-neous equation with variable coefficients, we can always find a second inde-pendent solution if one solution is known. This method is called reduction oforder.

Suppose u1(t) is a solution of the general equation

d2u

dt2+ k(t)

du

dt+ p(t)u = 0. (24)

Assume that u2(t) = v(t)u1(t) is a solution. We wish to find v(t) so that u2

is indeed a solution. However, v(t) must not be constant, as that would notsupply an independent solution. A straightforward substitution of u2 = vu1

into the differential equation leads to

v′′u1 + 2v′u′1 + vu′′

1 + k(t)(v′u1 + vu′1) + p(t)vu1 = 0.


Now collect terms in the derivatives of v. The preceding equation becomes

u1v′′ + (

2u′1 + k(t)u1

)v′ + (

u′′1 + k(t)u′

1 + p(t)u1

)v = 0.

However, u1 is a solution of Eq. (24), so the coefficient of v is 0. This leaves

u1v′′ + (

2u′1 + k(t)u1

)v′ = 0, (25)

which is a first-order linear equation for v′. Thus, a nonconstant v can befound, at least in terms of some integrals.

Example.Consider the equation(

1 − t2)u′′ − 2tu′ + 2u = 0, −1 < t < 1,

which has u1(t) = t as a solution. By assuming that u2 = v · t and substituting,we obtain (

1 − t2)(v′′t + 2v′) − 2t(v′t + v) + 2vt = 0.

After collecting terms, we have(1 − t2

)tv′′ + (2 − 4t2)v′ = 0.

From here, it is fairly easy to find

v′′

v′ = 4t2 − 2

t(1 − t2)= −2

t+ 1

1 − t− 1

1 + t

(using partial fractions), and then

lnv′ = −2 ln(t) − ln(1 − t) − ln(1 + t).

Finally, each side is exponentiated to obtain

v′ = 1

t2(1 − t2)= 1

t2+ 1/2

1 − t+ 1/2

1 + t,

v = −1

t+ 1

2ln

∣∣∣∣1 + t

1 − t

∣∣∣∣. �

D. Higher-Order EquationsLinear homogeneous equations of order higher than 2 — especially order 4 —occur frequently in elasticity and fluid mechanics. A general, nth-order homo-geneous linear equation may be written

u(n) + k1(t)u(n−1) + · · · + kn−1(t)u(1) + kn(t)u = 0, (26)


Root Multiplicity Contributionm (real) 1 cemt

m (real) k (c1 + c2t + · · · + cktk−1)emt

m,m (complex) 1 (a cos(βt) + b sin(βt))eαt

m = α + iβ

m,m (complex) k(a1 + a2t + · · · + aktk−1) cos(βt)eαt

+ (b1 + b2t + · · · + bktk−1) sin(βt)eαt

Table 3 Contributions to general solution

in which the coefficients k1(t), k2(t), etc., are given functions of t. The tech-niques of solution are analogous to those for second-order equations. In par-ticular, they depend on the Principle of Superposition, which remains validfor this equation. That principle allows us to say that the general solutionof Eq. (26) has the form of a linear combination of n independent solutionsu1(t),u2(t), . . . ,un(t) with arbitrary constant coefficients,

u(t) = c1u1(t) + c2u2(t) + · · · + cnun(t).

Of course, we cannot solve the general nth-order equation (26), but we canindeed solve any homogeneous linear equation with constant coefficients,

u(n) + k1u(n−1) + · · · + kn−1u(1) + knu = 0. (27)

We must now find n independent solutions of this equation. As in the second-order case, we assume that a solution has the form u(t) = emt , and find valuesof m for which this is true. That is, we substitute emt for u in the differen-tial equation (27) and divide out the common factor of emt . The result is thepolynomial equation

mn + k1mn−1 + · · · + kn−1m + kn = 0, (28)

called the characteristic equation of the differential equation (27).Each distinct root of the characteristic equation contributes as many inde-

pendent solutions as its multiplicity, which might be as high as n. Recall alsothat the polynomial equation (28) may have complex roots, which will occurin conjugate pairs if — as we assume — the coefficients k1, k2, etc., are real.When this happens, we prefer to have real solutions, in the form of an expo-nential times sine or cosine, instead of complex exponentials. The contributionof each root or pair of conjugate roots of Eq. (28) is summarized in Table 3.Since the sum of the multiplicities of the roots of Eq. (28) is n, the sum of thecontributions produces a solution with n terms, which can be shown to be thegeneral solution.


Example.Find the general solution of this fourth-order equation

u(4) + 3u(2) − 4u = 0.

The characteristic equation is m4 +3m2 −4 = 0, which is easy to solve becauseit is a biquadratic. We find that m2 = −4 or 1, and thus the roots are m = ±2i,±1, all with multiplicity 1. From Table 3 we find that a cos(2t)+ b sin(2t) cor-responds to the complex conjugate pair, m = ±2i, while et and e−t correspondto m = 1 and m = −1. Thus we build up the general solution,

u(t) = a cos(2t) + b sin(2t) + c1et + c2e−t . �

Example.Find the general solution of the fourth-order equation

u(4) − 2u(2) + u = 0.

The characteristic equation is m4 − 2m2 + 1 = 0, whose roots, found as in thepreceding, are ±1, both with multiplicity 2. From Table 3 we find that each ofthe roots contributes a first-degree polynomial times an exponential. Thus, weassemble the general solution as

u(t) = (c1 + c2t)et + (c3 + c4t)e−t .

With sinh(t) = (et − e−t)/2 and cosh(t) = (et + e−t)/2, the terms of the pre-ceding combination can be rearranged to give the general solution in a differ-ent form,

u(t) = (C1 + C2t) cosh(t) + (C3 + C4t) sinh(t). �

Some important equations and their solutions.

1.du

dt= ku (k is constant),

u(t) = cekt .

2.d2u

dt2+ λ2u = 0,

u(t) = a cos(λt) + b sin(λt).

3.d2u

dt2− λ2u = 0,

u(t) = a cosh(λt) + b sinh(λt) or u(t) = c1eλt + c2e−λt .

4. t2u′′ + tu′ − λ2u = 0,

u(t) = c1tλ + c2t−λ.


E X E R C I S E S

In Exercises 1–6, find the general solution of the differential equation. Be care-ful to identify the dependent and independent variables.

1.d2φ

dx2+ λ2φ = 0.

2.d2φ

dx2− µ2φ = 0.

3.d2u

dt2= 0.

4.dT

dt= −λ2kT.

5.1

r

d

dr

(r

dw

dr

)− λ2

r2w = 0.

6. ρ2 d2R

dρ2+ 2ρ

dR

dρ− n(n + 1)R = 0.

In Exercises 7–11, find the general solution. In some cases, it is helpful to carryout the indicated differentiation, in others it is not.

7.d

dx

((h + kx)

dv

dx

)= 0 (h, k are constants).

8. (exφ′)′ + λ2exφ = 0.

9.d

dx

(x3 du

dx

)= 0.

10. r2 d2u

dr2+ r

du

dr+ λ2u = 0.

11.1

r

d

dr

(r

du

dr

)= 0.

12. Compare and contrast the form of the solutions of these three differentialequations and their behavior as t → ∞.

a.d2u

dt2+ u = 0; b.

d2u

dt2= 0; c.

d2u

dt2− u = 0.

In Exercises 13–15, use the “exponential guess” method to find the generalsolution of the differential equations (λ is constant).

13.d4u

dx4− λ4u = 0.

14.d4u

dx4+ λ4u = 0.


15.d4u

dx4+ 2λ2 d2u

dx2+ λ4u = 0.

In Exercises 16–18, one solution of the differential equation is given. Find asecond independent solution.

16.d2u

dt2+ 2a

du

dt+ a2u = 0, u1(t) = e−at .

17. t2 d2u

dt2+ (1 − 2b)t

du

dt+ b2u = 0, u1(t) = tb.

18.d

dx

(x

du

dx

)+ 4x2 − 1

4xu = 0, u1(x) = cos(x)√

x.

In Exercises 19–21, use the indicated change of variable to solve the differentialequation.

19.d

dρ

(ρ2 dR

dρ

)+ λ2ρ2R = 0, R(ρ) = u(ρ)

ρ.

20.d

dρ

(ρ

dφ

dρ

)+ 4λ2ρ2 − 1

4ρφ = 0, φ(ρ) = v(ρ)√

ρ.

21. t2 d2u

dt2+ kt

du

dt+ pu = 0, x = ln t, u(t) = v(x).

22. Solve each initial value problem. Assuming that the solution representsthe displacement of a mass in a mass–spring–damper system, as in thetext, describe the motion in words.

a.d2u

dt2+ 4u = 0, u(0) = 1,

du

dt(0) = 0;

b.d2u

dt2+ 2

du

dt+ 2u = 0, u(0) = 1,

du

dt(0) = 1;

c.d2u

dt2+ 2

du

dt+ u = 0, u(0) = 1,

du

dt(0) = 1;

d.d2u

dt2+ 2

du

dt+ 0.75u = 0, u(0) = 0,

du

dt(0) = 1.

23. Sheet metal is produced by repeatedly feeding the sheet between steelrollers to reduce the thickness. In the article “On the characteristics andmechanism of rolling instability and chatter” [Y.-J. Lin et al., J. of Manu-facturing Science and Engineering, 125 (2003): 778–786], the authors findthat the distance between rollers is well approximated by h + y, where his the nominal output thickness and y is the solution of the differentialequation y′′ + 2αy′ + σ 2y = 0. The elasticity of the sheet and the rollersprovides the restoring force, and the plastic deformation of the sheet ef-fectively provides damping.


For high-speed operation, the system is underdamped. Solve the initialvalue problem consisting of the differential equation and the initial con-ditions y(0) = −0.001h, y′(0) = 0.

24. (Continuation) For an input speed of 25.4 m/s, it is observed that σ ∼=600 Hz or 1200π radians/s and α = 0.103σ . Using these values, obtaina graph of the solution of the preceding exercise, over the range 0 < t <

0.02 s. How far does the sheet move in 0.02 s?

25. (Continuation) The damping constant α referred to in the previous ex-ercises appears to depend on v, the speed of the sheet into the rollers,according to the relation α/σ = A/v, where A is a constant. From the in-formation given previously, the value of A is about 2.62. Assuming this iscorrect, find the speed v at which damping is critical.

0.2 Nonhomogeneous Linear EquationsIn this section, we will review methods for solving nonhomogeneous linearequations of first and second orders,

du

dt= k(t)u + f (t),

d2u

dt2+ k(t)

du

dt+ p(t)u = f (t).

Of course, we assume that the inhomogeneity f (t) is not identically 0. Thesimplest nonhomogeneous equation is

du

dt= f (t). (1)

This can be solved in complete generality by one integration:

u(t) =∫

f (t)dt + c. (2)

We have used an indefinite integral and have written c as a reminder that thereis an arbitrary additive constant in the general solution of Eq. (1). A moreprecise way to write the solution is

u(t) =∫ t

t0

f (z)dz + c. (3)

Here we have replaced the indefinite integral by a definite integral with vari-able upper limit. The lower limit of integration is usually an initial time. Note

0.2 Nonhomogeneous Linear Equations 15

that the name of the integration variable is changed from t to something else(here, z) to avoid confusing the limit with the dummy variable of integration.The simple second-order equation

d2u

dt2= f (t) (4)

can be solved by two successive integrations.The two theorems that follow summarize some properties of linear equa-

tions that are useful in constructing solutions.

Theorem 1. The general solution of a nonhomogeneous linear equation has theform u(t) = up(t) + uc(t), where up(t) is any particular solution of the nonho-mogeneous equation and uc(t) is the general solution of the corresponding homo-geneous equation. �

Theorem 2. If up1(t) and up2(t) are particular solutions of a differential equationwith inhomogeneities f1(t) and f2(t), respectively, then k1up1(t) + k2up2 is a par-ticular solution of the differential equation with inhomogeneity k1f1(t) + k2f2(t)(k1, k2 are constants). �

Example.Find the solution of the differential equation

d2u

dt2+ u = 1 − e−t .

The corresponding homogeneous equation is

d2u

dt2+ u = 0,

with general solution uc(t) = c1 cos(t) + c2 sin(t) (found in Section 1). A par-ticular solution of the equation with the inhomogeneity f1(t) = 1, that is, ofthe equation

d2u

dt2+ u = 1,

is up1(t) = 1. A particular solution of the equation

d2u

dt2+ u = e−t

is up2(t) = 12 e−t . (Later in this section, we will review methods for constructing

these particular solutions.) Then, by Theorem 2, a particular solution of thegiven nonhomogeneous Eq. (5) is up(t) = 1 − 1

2 e−t . Finally, by Theorem 1, the


Inhomogeneity, f (t) Form of Trial Solution, up(t)(a0tn + a1tn−1 + · · · + an)eαt (A0tn + A1tn−1 + · · · + An)eαt

(a0tn + · · · + an)eαt cos(βt)+ (b0tn + · · · + bn)eαt sin(βt)

(A0tn + · · · + An)eαt cos(βt)+ (B0tn + · · · + Bn)eαt sin(βt)

Table 4 Undetermined coefficients

general solution of the given equation is

u(t) = 1 − 1

2e−t + c1 cos(t) + c2 sin(t).

If two initial conditions are given, then c1 and c2 are available to satisfy them.Of course, an initial condition applies to the entire solution of the given dif-ferential equation, not just to uc(t). �

Now we turn our attention to methods for finding particular solutions ofnonhomogeneous linear differential equations.

A. Undetermined CoefficientsThis method involves guessing the form of a trial solution and then finding theappropriate coefficients. Naturally, it is limited to the cases in which we canguess successfully: when the equation has constant coefficients and the inho-mogeneity is simple in form. Table 4 offers a summary of admissible inhomo-geneities and the corresponding forms for particular solution. The parametersn, α,β and the coefficients a0, . . . ,an, b0, . . . ,bn are found by inspecting thegiven inhomogeneity. The table compresses several cases. For instance, f (t) inline 1 is a polynomial if α = 0 or an exponential if n = 0 and α �= 0. In line 2,both sine and cosine must be included in the trial solution even if one is absentfrom f (t); but α = 0 is allowed, and so is n = 0.

Example.Find a particular solution of

d2u

dt2+ 5u = te−t .

We use line 1 of Table 4. Evidently, n = 1 and α = −1. The appropriate formfor the trial solution is

up(t) = (A0t + A1)e−t .

When we substitute this form into the differential equation, we obtain

(A0t + A1 − 2A0)e−t + 5(A0t + A1)e−t = te−t .


Now, equating coefficients of like terms gives these two equations for the coef-ficients:

6A0 = 1 (coefficient of te−t),

6A1 − 2A0 = 0 (coefficient of e−t).

These we solve easily to find A0 = 1/6, A1 = 1/18. Finally, a particular solutionis

up(t) =(

1

6t + 1

18

)e−t . �

A trial solution from Table 4 will not work if it contains any term that is a so-lution of the homogeneous differential equation. In that case, the trial solutionhas to be revised by the following rule.

Revision Rule. Multiply by the lowest positive integral power of t such that noterm in the trial solution satisfies the corresponding homogeneous equation. �

Example.Table 4 suggests the trial solution up(t) = (A0t + A1)e−t for the differentialequation

d2u

dt2− u = te−t .

However, we know that the solution of the corresponding homogeneous equa-tion, u′′ − u = 0, is

uc(t) = c1 et + c2 e−t .

The trial solution contains a term (A1e−t) that is a solution of the homoge-neous equation. Multiplying the trial solution by t eliminates the problem.Thus, the trial solution is

up(t) = t(A0t + A1)e−t = (A0t2 + A1t

)e−t .

Similarly, the trial solution for the differential equation

d2u

dt2+ 2

du

dt+ u = te−t

has to be revised. The solution of the corresponding homogeneous equationis uc(t) = c1 e−t + c2 te−t . The trial solution from the table has to be multipliedby t2 to eliminate solutions of the homogeneous equation. �

Example: Forced Vibrations.The displacement u(t) of a mass in a mass–spring–damper system, starting


Figure 2 Mass–spring–damper system with an external force.

from rest, with an external sinusoidal force (see Fig. 2) is described by thisinitial value problem:

d2u

dt2+ b

du

dt+ ω2u = f0 cos(µt),

u(0) = 0,du

dt(0) = 0.

See the Section 1 example on the mass–spring–damper system. The coeffi-cient f0 is proportional to the magnitude of the force. There are three impor-tant cases.

b = 0,µ �= ω: undamped, no resonance. The form of the trial solution is

up(t) = A cos(µt) + B sin(µt).

Substitution and simple algebra lead to the particular solution

u0(t) = f0

ω2 − µ2cos(µt)

(that is, B = 0). The general solution of the differential equation is

u(t) = f0

ω2 − µ2cos(µt) + c1 cos(ωt) + c2 sin(ωt).

Applying the initial conditions determines c1 and c2. Finally, the solution ofthe initial value problem is

u(t) = f0

ω2 − µ2

(cos(µt) − cos(ωt)

).


b = 0, µ = ω: resonance. Now, since ω = µ, the trial solution must berevised to

up(t) = At cos(µt) + Bt sin(µt).

Substitution into the differential equation and simple algebra give A = 0, B =f0/2µ, or

up(t) = f0

2µt sin(µt).

The general solution of the differential equation is

u(t) = f0

2µt sin(µt) + c1 cos(µt) + c2 sin(µt).

(Remember that b = 0 and ω = µ.) The initial conditions give c1 = c2 = 0, sothe solution of the initial value problem is

u(t) = f0

2µt sin(µt).

The presence of the multiplier t means that the amplitude of the oscillation isincreasing. This is the phenomenon of resonance.

b > 0: damped motion. The ideas are straightforward applications of thetechniques developed earlier. The trial solution is a combination of cos(µt)and sin(µt). Somewhat less simple algebra gives

up(t) = f0

((ω2 − µ2) cos(µt) + µb sin(µt)

),

where = (ω2 −µ2)2 +µ2b2. The general solution of the differential equationmay take different forms, depending on the relation between b and ω. (SeeSection 1.) Assuming the underdamped case holds, we have

u(t) = f0

((ω2 − µ2) cos(µt) + µb sin(µt)

)+ e−bt/2

(c1 cos(γ t) + c2 sin(γ t)

)for the general solution of the differential equation. Here, γ = √

ω2 − (b/2)2

is real because we assumed underdamping.Applying the initial conditions gives, after some nasty algebra,

c1 = − f0

(ω2 − µ2

), c2 = − f0

b

γ

ω2 + µ2

2.


Notice that, as t increases, the terms that come from the complementarysolution approach 0, while the terms that come from the particular solutionpersist. These cases are illustrated with animation on the CD. �

B. Variation of ParametersGenerally, if a linear homogeneous differential equation can be solved, the cor-responding nonhomogeneous equation can also be solved, at least in terms ofintegrals.

1. First-order equationsSuppose that uc(t) is a solution of the homogeneous equation

du

dt= k(t)u. (5)

Then to find a particular solution of the nonhomogeneous equation

du

dt= k(t)u + f (t), (6)

we assume that up(t) = v(t)uc(t). Substituting up in this form into the differ-ential equation (6) we have

dv

dtuc + v

duc

dt= k(t)vuc + f (t). (7)

However, u′c = k(t)uc, so one term on the left cancels a term on the right,

leaving

dv

dtuc = f (t), or

dv

dt= f (t)

uc(t). (8)

The latter is a nonhomogeneous equation of simplest type, which can besolved for v(t) in one integration.

Example.Use this method to find a solution of the nonhomogeneous equation

du

dt= 5u + t.

We should try the form up(t) = v(t) · e5t , because e5t is a solution of u′ = 5u.Substituting the preceding form for up, we find

dv

dt· e5t + v · 5e5t = 5ve5t + t,


or, after canceling 5ve5t from both sides and simplifying, we find

dv

dt= e−5t t.

This equation is integrated once (by parts) to find

v(t) =(

− t

5− 1

25

)e−5t .

From here, we obtain up(t) = v(t) · e5t = −(15 t + 1

25

). �

2. Second-order equationsTo find a particular solution of the nonhomogeneous second-order equation

d2u

dt2+ k(t)

du

dt+ p(t)u = f (t), (9)

we need two independent solutions, u1(t) and u2(t), of the corresponding ho-mogeneous equation

d2u

dt2+ k(t)

du

dt+ p(t)u = 0. (10)

Then we assume that our particular solution has the form

up(t) = v1(t)u1(t) + v2(t)u2(t), (11)

where v1 and v2 are functions to be found. If we simply insert up in this forminto Eq. (9), we obtain one complicated second-order equation in two un-known functions. However, if we impose the extra requirement that

dv1

dtu1 + dv2

dtu2 = 0, (12)

then we find that

u′p = v′

1u1 + v′2u2 + v1u′

1 + v2u′2 = v1u′

1 + v2u′2, (13)

u′′p = v′

1u′1 + v′

2u′2 + v1u′′

1 + v2u′′2, (14)

and the equation that results from substituting Eq. (11) into Eq. (9) becomes

v′1u′

1 + v′2u′

2 + v1(u′′1 + k(t)u′

1 + p(t)u1) + v2(u′′2 + k(t)u′

2 + p(t)u2) = f (t).

This simplifies further: The multipliers of v1 and v2 are both 0, because u1 andu2 satisfy the homogeneous Eq. (10).


Thus, we are left with a pair of simultaneous equations,

v′1u1 + v′

2u2 = 0, (12′)

v′1u′

1 + v′2u′

2 = f (t), (15)

in the unknowns v′1 and v′

2. The determinant of this system is∣∣∣∣∣u1 u2

u′1 u′

2

∣∣∣∣∣ = W(t), (16)

the Wronskian of u1 and u2. Since these were to be independent solutions ofEq. (10), their Wronskian is nonzero, and we may solve for v′

1(t) and v′2(t) and

hence for v1 and v2.

Example.Use variation of parameters to solve the nonhomogeneous equation

d2u

dt2+ u = cos(ωt).

Assume a solution in the form

up(t) = v1 cos(t) + v2 sin(t),

because sin(t) and cos(t) are independent solutions of the corresponding ho-mogeneous equation u′′ + u = 0. The assumption of Eq. (12) is

v′1 cos(t) + v′

2 sin(t) = 0. (17)

Then our equation reduces to the following, corresponding to Eq. (15):

−v′1 sin(t) + v′

2 cos(t) = cos(ωt). (18)

Now we solve Eqs. (17) and (18) simultaneously to find

v′1 = − sin(t) cos(ωt), v′

2 = cos(t) cos(ωt). (19)

These equations are to be integrated to find v1 and v2, and then up(t). �

Finally, we note that v1(t) and v2(t) can be found from Eqs. (12) and (15)in general:

v′1 = −u2f

W, v′

2 = u1f

W. (20)

Integrating these two equations, we find that

v1(t) = −∫

u2(t)f (t)

W(t)dt, v2(t) =

∫u1(t)f (t)

W(t)dt. (21)


Now, Eq. (11) may be used to form a particular solution of the nonhomoge-neous equation (9).

We may also obtain v1 and v2 by using definite integrals with variable upperlimit:

v1(t) = −∫ t

t0

u2(z)f (z)

W(z)dz, v2(t) =

∫ t

t0

u1(z)f (z)

W(z)dz. (22)

The lower limit is usually the initial value of t, but may be any convenientvalue. The particular solution can now be written as

up(t) = −u1(t)

∫ t

t0

u2(z)f (z)

W(z)dz + u2(t)

∫ t

t0

u1(z)f (z)

W(z)dz.

Furthermore, the factors u1(t) and u2(t) can be inside the integrals (which arenot with respect to t), and these can be combined to give a tidy formula, asfollows.

Theorem 3. Let u1(t) and u2(t) be independent solutions of

d2u

dt2+ k(t)

du

dt+ p(t)u = 0 (H)

with Wronskian W(t) = u1(t)u′2(t) − u2(t)u′

1(t). Then

up(t) =∫ t

t0

G(t, z)f (z)dz

is a particular solution of the nonhomogeneous equation

d2u

dt2+ k(t)

du

dt+ p(t)u = f (t), (NH)

where G is the Green’s function defined by

G(t, z) = u1(z)u2(t) − u2(z)u1(t)

W(z). (23)

�

E X E R C I S E S

In Exercises 1–10, find the general solution of the differential equation.


1.du

dt+ a(u − T) = 0.

3.du

dt+ au = e−at .

5.d2u

dt2+ u = cos(t).

7.d2u

dt2+ 3

du

dt+ 2u = cosh(t).

9.1

ρ2

d

dρ

(ρ2 du

dρ

)= −1.

2.du

dt+ au = eat .

4.d2u

dt2+ u = cos(ωt) (ω �= 1).

6.d2u

dx2− γ 2(u − U) = 0

(U, γ 2 are constants).

8.1

r

d

dr

(r

du

dr

)= −1.

10.d2u

dt2= −1.

11. Let h(t) be the height of a parachutist above the surface of the earth. Con-sideration of forces on his body leads to the initial value problem for h:

Md2h

dt2+ K

dh

dt= −Mg,

h(0) = h0,dh

dt(0) = 0

(M = mass, g = acceleration of gravity, K = parachute constant). Solvethe problem, taking g = 32 ft/s2 and K/M = 0.1/s.

12. Solve this initial value problem for forced vibrations,

d2u

dt2+ ω2u = f0 sin(µt),

u(0) = 0,du

dt(0) = 0,

in two cases: (a) µ �= ω, (b) µ = ω.In Exercises 13–19, use variation of parameters to find a particular solution ofthe differential equation. Be sure that the differential equation is in the correctform.

13.du

dt+ au = e−at , uc(t) = e−at .

14. tdu

dt= −1, uc(t) = 1.

15.d2y

dx2+ y = tan(x), y1(x) = cos(x), y2(x) = sin(x).

16.d2y

dx2+ y = sin(x), y1(x) = cos(x), y2(x) = sin(x).


17.d2u

dt2= −1, u1(t) = 1, u2(t) = t.

18.1

r

d

dr

(r

du

dr

)= −1, u1(r) = 1, u2(r) = ln(r).

19. t2 d2u

dt2+ t

du

dt− u = 1, u1(t) = t, u2(t) = 1

t.

In Exercises 20–22, use Theorem 3 to develop the formula shown for a partic-ular solution of the differential equation.

20.d2u

dt2+ γ 2u = f (t), up(t) = 1

γ

∫ t

0sin

(γ (t − z)

)f (z)dz.

21.du

dt+ au = f (t), up(t) =

∫ t

0e−a(t−z)f (z)dz.

22.d2u

dt2− γ 2u = f (t), up(t) = 1

γ

∫ t

0sinh

(γ (t − z)

)f (z)dz.

23. In “Model for temperature estimation of electric couplings suffering heavylightning currents” [A.D. Polykriti et al., IEE Proceedings — Generation,Transmission and Distribution, 151 (2004): 90–94], the authors model thetemperature rise above ambient in a coupling with this initial value prob-lem:

ρcdT

dt= i2(t)R(1 + αT), T(0) = 0.

Parameters: ρ is density, c is specific heat, i(t) is the current due to a light-ening strike, R is the resistance of the coupling at ambient temperature,and the factor (1 + αT) shows how resistance increases with temperature.

Simplify the differential equation algebraically to get

dT

dt= Ki2(t)(β + T), T(0) = 0,

and identify β and K in terms of the other parameters.

24. (Continuation) The authors model the lightning current with the func-tion i(t) = Imax(e−λt − e−µt)/n, where n is a factor to make Imax the ac-tual maximum. Obtain graphs of this function and the simpler functioni(t) = Imaxe−λt , using these values: Imax = 100 kA, n = 0.93, λ = 2.1,µ = 150. The unit for time is milliseconds. Graph for t from 0 to 2 ms,which is the range of interest.

25. (Continuation) Solve the initial value problem using the simpler functionfor current. (Don’t forget to square.) Graph the result for t from 0 to 2 ms,using β = 0.26 and K = 13.


0.3 Boundary Value ProblemsA boundary value problem in one dimension is an ordinary differential equa-tion together with conditions involving values of the solution and/or its deriv-atives at two or more points. The number of conditions imposed is equal tothe order of the differential equation. Usually, boundary value problems of anyphysical relevance have these characteristics: (1) The conditions are imposedat two different points; (2) the solution is of interest only between those twopoints; and (3) the independent variable is a space variable, which we shallrepresent as x. In addition, we are primarily concerned with cases where thedifferential equation is linear and of second order. However, problems in elas-ticity often involve fourth-order equations.

In contrast to initial value problems, even the most innocent lookingboundary value problem may have exactly one solution, no solution, or aninfinite number of solutions. Exercise 1 illustrates these cases.

When the differential equation in a boundary value problem has a knowngeneral solution, we use the two boundary conditions to supply two equationsthat are to be satisfied by the two constants in the general solution. If the dif-ferential equation is linear, these are two linear equations and can be easilysolved, if there is a solution.

In the rest of this section we examine some physical examples that are natu-rally associated with boundary value problems.

Example: Hanging Cable.First we consider the problem of finding the shape of a cable that is fastenedat each end and carries a distributed load. The cables of a suspension bridgeprovide an important example. Let u(x) denote the position of the centerlineof the cable, measured upward from the x-axis, which we assume to be hori-zontal. (See Fig. 3.) Our objective is to find the function u(x).

The shape of the cable is determined by the forces acting on it. In our analy-sis, we consider the forces that hold a small segment of the cable in place. (SeeFig. 4.) The key assumption is that the cable is perfectly flexible. This meansthat force inside the cable is always a tension and that its direction at everypoint is the direction tangent to the centerline.

Figure 3 The hanging cable.

0.3 Boundary Value Problems 27

We suppose that the cable is not moving. Then by Newton’s second law,the sum of the horizontal components of the forces on the segment is 0, andlikewise for the vertical components. If T(x) and T(x+x) are the magnitudesof the tensions at the ends on the segment, we have these two equations:

T(x + x) cos(φ(x + x)

) − T(x) cos(φ(x)

) = 0 (Horizontal), (1)

T(x + x) sin(φ(x + x)

) − T(x) sin(φ(x)

) − f (x)x = 0 (Vertical). (2)

In the second equation, f (x) is the intensity of the distributed load, measuredin force per unit of horizontal length, so f (x)x is the load borne by the smallsegment.

From Eq. (1) we see that the horizontal component of the tension is the sameat both ends of the segment. In fact, the horizontal component of tension hasthe same value — call it T — at every point, including the endpoints wherethe cable is attached to solid supports. By simple algebra we can now find thetension in the cable at the ends of our segment,

T(x + x) = T

cos(φ(x + x)

) , T(x) = T

cos(φ(x)

) ,

and substitute these into Eq. (2), which becomes

T

cos(φ(x + x)

) sin(φ(x + x)

) − T

cos(φ(x)

) sin(φ(x)

) − f (x)x = 0

or

T(tan

(φ(x + x)

) − tan(φ(x)

)) − f (x)x = 0.

Before going further we should note (Fig. 4) that φ(x) measures the anglebetween the tangent to the centerline of the cable and the horizontal. As theposition of the centerline is given by u(x), tan(φ(x)) is just the slope of thecable at x. From elementary calculus we know

tan(φ(x)

) = du

dx(x).

Figure 4 Section of cable showing forces acting on it. The angles are α = φ(x),β = φ(x + x).


Substituting the derivative for the slope and making some algebraic adjust-ments, we obtain

T(u′(x + x) − u′(x)

) = f (x)x.

Dividing through by x yields

Tu′(x + x) − u′(x)

x= f (x).

In the limit, as x approaches 0, the difference quotient in the left memberbecomes the second derivative of u, and the result is the equation

Td2u

dx2= f (x), (3)

which is valid for x in the range 0 < x < a, where the cable is located. In addi-tion, u(x) must satisfy the boundary conditions

u(0) = h0, u(a) = h1. (4)

For any particular case, we must choose an appropriate model for the load-ing, f (x). One possibility is that the cable is hanging under its own weight ofw units of weight per unit length of cable. Then in Eq. (2), we should put

f (x)x = ws

xx,

where s represents arc length along the cable. In the limit, as x approaches 0,s/x has the limit

limx→0

s

x=

√1 +

(du

dx

)2

.

Therefore, with this assumption, the boundary value problem that determinesthe shape of the cable is

d2u

dx2= w

T

√1 +

(du

dx

)2

, 0 < x < a, (5)

u(0) = h0, u(a) = h1. (6)

Notice that the differential equation is nonlinear. Nevertheless, we can find itsgeneral solution in closed form and satisfy the boundary conditions by appro-priate choice of the arbitrary constants that appear. (See Exercises 4 and 5.)

Another case arises when the cable supports a load uniformly distributed inthe horizontal direction, as given by

f (x)x = wx.


Figure 5 Cylinder of heat-conducting material.

This is approximately true for a suspension bridge. The boundary value prob-lem to be solved is then

d2u

dx2= w

T, 0 < x < a,

u(0) = h0, u(a) = h1. (7)

The general solution of the differential equation (7) can be found by theprocedures of Sections 1 and 2. It is

u(x) =(

w

2T

)x2 + c1x + c2,

where c1 and c2 are arbitrary. The two boundary conditions require

u(0) = h0: c2 = h0,

u(a) = h1:

(w

2T

)a2 + c1a + c2 = h1.

These two are solved for c1 and c2 in terms of given parameters. The result,after some beautifying algebra, is

u(x) = w

2T

(x2 − ax

) + h1 − h0

ax + h0. (8)

Clearly, this function specifies the cable’s shape as part of a parabola openingupward. �

Example: Heat Conduction in a Rod.A long rod of uniform material and cross section conducts heat along its axialdirection (see Fig. 5). We assume that the temperature in the rod, u(x), doesnot change in time. A heat balance (“what goes in must come out”) applied toa slice of the rod between x and x + x (Fig. 6) shows that the heat flow rateq, measured in units of heat per unit time per unit area, obeys the equation

q(x)A + g(x)Ax = q(x + x)A, (9)


Figure 6 Section cut from heat-conducting cylinder showing heat flow.

in which A is the cross-sectional area and g is the rate at which heat enters theslice by means other than conduction through the two faces. For instance, ifheat is generated in the slice by an electric current I, we might have

g(x)Ax = I2Rx, (10)

where R is the resistance of the rod per unit length. If heat is lost throughthe cylindrical surface of the rod by convection to a surrounding medium attemperature T, then g(x) would be given by “Newton’s law of cooling,”

g(x)Ax = −h(u(x) − T)C x, (11)

where C is the circumference of the rod and h is the heat transfer coefficient.(This minus sign appears because, if u(x) > T, heat actually leaves the rod.)

Equation (9) may be altered algebraically to read

q(x + x) − q(x)

x= g(x),

and application of the limiting process leaves

dq

dx= g(x). (12)

The unknown function u(x) does not appear in Eq. (12). However, a well-known experimental law (Fourier’s law) says that the heat flow rate through aunit area of material is directly proportional to the temperature difference andinversely proportional to thickness. In the limit, this law takes the form

q = −κdu

dx. (13)

The minus sign expresses the fact that heat moves from hotter toward coolerregions.

Combining Eqs. (12) and (13) gives the differential equation

−κd2u

dx2= g(x), 0 < x < a, (14)

where a is the length of the rod and the conductivity κ is assumed to be con-stant.


If the two ends of the rod are held at constant temperature, the boundaryconditions on u would be

u(0) = T0, u(a) = T1. (15)

On the other hand, if heat were supplied at x = 0 (by a heating coil, for in-stance), the boundary condition there would be

−κAdu

dx(0) = H, (16)

where H is measured in units of heat per unit time.

Example.Solve the problem

−κd2u

dx2= −hu(x)

C

A, 0 < x < a, (17)

u(0) = T0, u(a) = T0. (18)

(Physically, the rod is losing heat to a surrounding medium at temperature 0,while both ends are held at the same temperature T0.) If we designate µ2 =hC/κA, the differential equation becomes

d2u

dx2− µ2u = 0, 0 < x < a,

with general solution

u(x) = c1 cosh(µx) + c2 sinh(µx).

Application of the boundary condition at x = 0 gives c1 = T0; the secondboundary condition requires that

u(a) = T0: T0 = T0 cosh(µa) + c2 sinh(µa).

Thus c2 = T0(1 − cosh(µa))/ sinh(µa) and

u(x) = T0

(cosh(µx) + 1 − cosh(µa)

sinh(µa)sinh(µx)

). �

It should be clear now that solving a boundary value problem is not sub-stantially different from solving an initial value problem. The procedure is (1)find the general solution of the differential equation, which must contain somearbitrary constants, and (2) apply the boundary conditions to determine val-ues for the arbitrary constants. In our examples the differential equations have


Figure 7 Column carrying load P.

Figure 8 Section of column showing forces and moments.

been of second order, causing the appearance of two arbitrary constants, whichare to be determined by the boundary conditions.

The next example is somewhat different in spirit from the others. Instead ofjust finding the solution of a boundary value problem, we will be looking forparameter values that permit the existence of solutions of special form.

Example: Buckling of a Column.A long, slender column whose bottom end is hinged carries an axial load asshown in Fig. 7. The upper end of the column can move up or down but notsideways. The displacement of the column’s centerline from a vertical referenceline is given by u(x). If the column were cut at any point x, an upward force Pand a clockwise moment Pu(x) would have to be applied to the upper part tokeep it in equilibrium (see Fig. 8). This force and moment must be suppliedby the lower part of the column.


It is known that the internal bending moment (positive when counterclock-wise) in a column is given by the product

EId2u

dx2,

where E is Young’s modulus and I is the moment of inertia of the cross-sectional area. (The moment I = b4/12 for a column whose cross section isa square of side b.) Thus equating the external moment to the internal mo-ment gives the differential equation

EId2u

dx2= −Pu, 0 < x < a, (19)

which, together with the boundary conditions

u(0) = 0, u(a) = 0, (20)

determines the function u(x).In order to study this problem more conveniently, we set

P

EI= λ2

so that the differential equation becomes

d2u

dx2+ λ2u = 0, 0 < x < a. (21)

Now, the general solution of this differential equation is

u(x) = c1 cos(λx) + c2 sin(λx).

As u(0) = 0, we must choose c1 = 0, leaving u(x) = c2 sin(λx). The secondboundary condition requires that

u(a) = 0: c2 sin(λa) = 0.

If sin(λa) is not 0, the only possibility is that c2 = 0. In this case we find thatthe solution is

u(x) ≡ 0, 0 < x < a.

Physically, this means that the column stands straight and transmits the loadto its support, as it was probably intended to do.

Something quite different happens if sin(λa) = 0, for then any choice ofc2 gives a solution. The physical manifestation of this case is that the columnassumes a sinusoidal shape and may then collapse, or buckle, under the axialload. Mathematically, the condition sin(λa) = 0 means that λa is an integer


multiple of π , since sin(π) = 0, sin(2π) = 0, etc., and integer multiples of π

are the only arguments for which the sine function is 0. The equation λa = π ,in terms of the original parameters, is√

P

EIa = π.

It is reasonable to think of E, I, and a as given quantities; thus it is the force

P = EI

(π

a

)2

,

called the critical or Euler load, that causes the buckling. The higher criticalloads, corresponding to λa = 2π , λa = 3π , etc., are so unstable as to be of nophysical interest in this problem. �

The buckling example is one instance of an eigenvalue problem. The gen-eral setting is a homogeneous differential equation containing a parameter λ

and accompanied by homogeneous boundary conditions. Because both dif-ferential equations and boundary conditions are homogeneous, the constantfunction 0 is always a solution. The question to be answered is: What valuesof the parameter λ allow the existence of nonzero solutions? Eigenvalue prob-lems often are employed to find the dividing line between stable and unstablebehavior. We will see them frequently in later chapters.

E X E R C I S E S

1. Of these three boundary value problems, one has no solution, one hasexactly one solution, and one has an infinite number of solutions. Whichis which?

a.d2u

dx2+ u = 0, u(0) = 0, u(π) = 0;

b.d2u

dx2+ u = 1, u(0) = 0, u(1) = 0;

c.d2u

dx2+ u = 0, u(0) = 0, u(π) = 1.

2. Find the Euler buckling load of a steel column with a 2 in. × 3 in. rectan-gular cross section. The parameters are E = 30 × 106 lb/in.2, I = 2 in.4,a = 10 ft.

3. Find all values of the parameter λ for which these homogeneous boundaryvalue problems have a solution other than u(x) ≡ 0.

a.d2u

dx2+ λ2u = 0, u(0) = 0,

du

dx(a) = 0;


b.d2u

dx2+ λ2u = 0,

du

dx(0) = 0, u(a) = 0;

c.d2u

dx2+ λ2u = 0,

du

dx(0) = 0,

du

dx(a) = 0.

4. Verify, by differentiating and substituting, that

u(x) = c′ + 1

µcosh

(µ(x + c)

)is the general solution of the differential equation (5). (Here µ = w/T.The graph of u(x) is called a catenary.)

5. Find the values of c and c′ for which the function u(x) in Exercise 4 satisfiesthe conditions

u(0) = h, u(a) = h.

6. A beam that is simply supported at its ends carries a distributed lateralload of uniform intensity w (force/length) and an axial tension load T(force). The displacement u(x) of its centerline (positive down) satisfiesthe boundary value problem here. Find u(x).

d2u

dx2− T

EIu = − w

EI

Lx − x2

2, 0 < x < L,

u(0) = 0, u(L) = 0.

7. The temperature u(x) in a cooling fin satisfies the differential equation

d2u

dx2= hC

κA(u − T), 0 < x < a,

and boundary conditions

u(0) = T0, −κdu

dx(a) = h

(u(a) − T

).

That is, the temperature at the left end is held at T0 > T while the surfaceof the rod and its right end exchange heat with a surrounding medium attemperature T. Find u(x).

8. Calculate the limit as a tends to infinity of u(x), the solution of the prob-lem in Exercise 7. Is the result physically reasonable?

9. In an electrical heating element, the temperature u(x) satisfies the bound-ary value problem that follows. Find u(x).

d2u

dx2= hC

κA(u − T) − I2R

κA, 0 < x < a,

u(0) = T, u(a) = T.


Figure 9 Poiseuille flow.

10. Verify that the solution of the problem given in Eqs. (17) and (18) can alsobe written as follows, with µ2 = Ch/Aκ :

u(x) = T0cosh(µ(x − a/2))

cosh(µa/2).

11. (Poiseuille flow) A viscous fluid flows steadily between two large paral-lel plates so that its velocity is parallel to the x-axis. (See Fig. 9.) Thex-component of velocity of the fluid at any point (x, y) is a function ofy only. It can be shown that this component u(x) satisfies the differentialequation

d2u

dy2= − g

µ, 0 < y < L,

where µ is the viscosity and −g is a constant, negative pressure gradi-ent. Find u(y), subject to the “no-slip” boundary conditions, u(0) = 0,u(L) = 0.

12. If the beam mentioned in Exercise 6 is subjected to axial compression in-stead of tension, the boundary value problem for u(x) becomes the onehere. Solve for u(x).

d2u

dx2+ P

EIu = − w

EI

Lx − x2

2, 0 < x < L,

u(0) = 0, u(L) = 0.

13. For what value(s) of the compressive load P in Exercise 12 does the prob-lem have no solution or infinitely many solutions?

14. The pressure p(x) in the lubricant under a plane pad bearing satisfies theproblem

d

dx

(x3 dp

dx

)= −K, a < x < b,

p(a) = 0, p(b) = 0.


Find p(x) in terms of a, b, and K (constant). Hint: The differential equa-tion can be solved by integration.

15. In a nuclear fuel rod, nuclear reaction constantly generates heat. If we treata rod as a one-dimensional object, the temperature u(x) in the rod mightsatisfy the boundary value problem

d2u

dx2+ g

κ= hC

κA(u − T), 0 < x < a,

u(0) = T, u(a) = T.

Here, g is the heat generation rate or power density, and the terms on theright-hand side represent heat transfer by convection to a surroundingmedium, usually pressurized water. Find u(x).

16. Sketch the solution of Exercise 15 and determine the maximum temper-ature encountered. Typical values for the parameters are g = 300 W/cm3,T = 325◦C, κ = 0.01 cal/cm s ◦C, a = 2.9 m, C/A = 4/cm, h =0.035 cal/cm2 s ◦C. It will be useful to know that 1 W = 0.239 cal/s.

17. An assembly of nuclear fuel rods is housed in a pressure vessel shapedroughly like a cylinder with flat or hemispherical ends. The temperaturein the thick steel wall of the vessel affects its strength and thus must bestudied for design and safety. Treating the vessel as a long cylinder (thatis, ignoring the effects of the ends), it is easy to derive this differentialequation in cylindrical coordinates for the temperature u(r) in the wall:

1

r

d

dr

(r du)

dr= 0, a < r < b,

where a and b are the inner and outer radii, respectively. The boundaryconditions both involve convection, with hot pressurized water at the in-ner radius and with air at the outer radius:

−κu′(a) = h0

(Tw − u(a)

),

κu′(b) = h1

(Ta − u(b)

).

Find u(r) in terms of the parameters, carefully checking the dimensions.

18. If a beam of uniform cross section is simply supported at its ends andcarries a distributed load w(x) along its length, then the displacement u(x)of its centerline satisfies the boundary value problem

d4u

dx4= w(x)

EI, 0 < x < a,

u(0) = 0, u′′(0) = 0, u(a) = 0, u′′(a) = 0.


(Here, E is Young’s modulus and I is the second moment of the cross sec-tion.) Solve this problem if w(x) = w0, constant.

19. If the beam of Exercise 18 is built into a wall at the left end and is unsup-ported at the right end, the boundary conditions become

u(0) = 0, u′(0) = 0, u′′(a) = 0, u′′′(a) = 0.

Solve the same differential equation subject to these conditions.

0.4 Singular Boundary Value ProblemsA boundary value problem can be singular in two different ways. In one case,an endpoint of the interval of interest is a singular point of the differentialequation. In the other, the interval is infinitely long.

Regular Singular PointRecall that a point x0 is a (regular) singular point of the differential equation

u′′ + k(x)u′ + p(x)u = f (x)

if the products

(x − x0)k(x), (x − x0)2p(x)

both have Taylor series expansions centered at x0 but either k(x) or p(x) orboth become infinite as x → x0. For example, the point x0 = 1 is a regularsingular point of the differential equation

(1 − x)u′′ + u′ + xu = 0.

In standard form, the equation is

u′′ + 1

1 − xu′ + x

1 − xu = 0.

Since both

k(x) = 1

1 − xand p(x) = x

1 − x

become infinite at x = 1, but (x − 1)k(x) and (x − 1)2p(x) both have Taylorseries expansions about the center x = 1, the point x0 = 1 is a regular singularpoint. Another convenient example is provided by the Cauchy–Euler equationof Section 1, which has a regular singular point at the origin.

This situation typically arises when a boundary point is a mathematicalboundary without being a physical boundary. For instance, a circular disk of

0.4 Singular Boundary Value Problems 39

radius c may be described in polar (r, θ) coordinates as occupying the region0 ≤ r ≤ c. The origin, at r = 0, is a mathematical boundary, yet physically thispoint is in the interior of the disk.

At a singular point, one cannot specify a value for u(x0), the solution ofthe differential equation, or for its derivative. However, it is usually necessaryto require that both u(x0) and u′(x0) be finite, or bounded. Tacitly, we alwaysrequire that the solution and its derivative be finite at every point of the intervalwhere we are solving a differential equation. But when a singular point is aboundary point of that interval, we enforce the condition explicitly. In theexample that follows we shall see how these conditions act so as to make thesolution of a boundary value problem unique.

Example: Radial Heat Flow.Suppose a long cylindrical bar, surrounded by a medium at temperature T,carries an electrical current. If heat flows in the radial direction much fasterthan in the axial direction, the temperature u(r) in the rod may be describedby the problem

1

r

d

dr

(r

du

dr

)= −H, 0 ≤ r < c, (1)

u(c) = T. (2)

Here, c is the radius of the rod, r is a polar coordinate, and H (constant) isproportional to the electrical power being converted into heat.

In this problem, only the physical boundary condition has been noted. Themathematical boundary r = 0 is a singular point, as is clear from the differen-tial equation in the form

d2u

dr2+ 1

r

du

dr= −H.

Thus, at this point we will require that u and du/dr be finite:

u(0), u′(0) finite. (3)

Now the differential equation (1) is easy to solve. Multiply through by r andintegrate once to find that

rdu

dr= −H

r2

2+ c1.

Divide through this equation by r and integrate once more to determine that

u(r) = −Hr2

4+ c1 ln(r) + c2.


Application of the special condition, that u(0) and u′(0) be finite, immediatelytells us that c1 = 0; for both, ln(r) and its derivative 1/r become infinite as rapproaches 0.

The physical boundary condition, Eq. (2), says that

u(c) = −Hc2

4+ c2 = T.

Hence, c2 = Hc2/4 + T, and the complete solution is

u(r) = H(c2 − r2)

4+ T. (4)

�

From this example, it is clear that the “artificial” boundary condition,boundedness of u(r) at the singular point r = 0, works just the way an or-dinary boundary condition works at an ordinary (not singular) point. It givesone condition to be fulfilled by the unknown constants c1 and c2, which arethen completely determined by the second boundary condition.

Semi-Infinite and Infinite IntervalsAnother type of singular boundary value problem is one for which the inter-val of interest is infinite. (Of course, this is always a mathematical abstractionthat cannot be realized physically.) For instance, on the interval 0 < x < ∞,sometimes called a semi-infinite interval, as it does have one finite endpoint, aboundary condition would normally be imposed at x = 0. At the other “end,”no boundary condition is imposed, because no boundary exists. However, wenormally require that both u(x) and u′(x) remain bounded as x increases. Inprecise terms, we require that there exist constants M and M′ for which∣∣u(x)

∣∣ ≤ M and∣∣u′(x)

∣∣ ≤ M′

are both satisfied for all x, no matter how large. We never identify M or M′,and the entire condition is usually written

u(x) and u′(x) bounded as x → ∞.

Example: Cooling Fin.A long cooling fin has one end held at a constant temperature T0 and ex-changes heat with a medium at temperature T through convection. The tem-perature u(x) in the fin satisfies the requirements

d2u

dx2= hC

κA(u − T), 0 < x, (5)

u(0) = T0 (6)

0.4 Singular Boundary Value Problems 41

(see Section 3). As the problem has been posed for a semi-infinite interval(because the fin is very long and, perhaps, to mask our ignorance of what ishappening at the other physical end), we must also impose the condition

u(x), u′(x) bounded as x → ∞. (7)

Now, the general solution of the differential equation (5) is

u(x) = T + c1 cosh(µx) + c2 sinh(µx),

where µ = √hC/κA. The boundary condition at x = 0 requires that

u(0) = T0: T + c1 = T0.

The boundedness condition, Eq. (7), requires that

c2 = −c1.

The reason for this is that of all the linear combinations of cosh and sinh, theonly one that is bounded as x → ∞ is

cosh(µx) − sinh(µx) = e−µx,

and its constant multiples. The final solution is easily found to be

u(x) = T + (T0 − T)(cosh(µx) − sinh(µx)

). �

Satisfying the boundedness condition in the example would have been sim-pler if we had expressed the general solution of the differential equation (5)as

u(x) = T + c′1eµx + c′

2e−µx.

We would have seen immediately that choosing c′1 = 0 is the only way to satisfy

the boundedness condition. We summarize the observation as a rule of thumb:the solution of

d2u

dx2− µ2 u = 0

on an interval I is best expressed as

u(x) ={

c1 cosh(µx) + c2 sinh(µx), if I is finite,

c1eµx + c2e−µx, if I is infinite.

E X E R C I S E S

1. Put each of the following equations in the form

u′′ + ku′ + pu = f


and identify the singular point(s).

a.1

r

d

dr

(r

du

dr

)= u;

c.d

dφ

(sin(φ)

du

dφ

)= sin(φ)u;

b.d

dx

((1 − x2

)du

dx

)= 0;

d.1

ρ2

d

dρ

(ρ2 du

dρ

)= −λ2u.

2. The temperature u in a large object having a hole of radius c in the middlemay be said to obey the equations

1

r

d

dr

(r

du

dr

)= 0, r > c,

u(c) = T.

Solve the problem, adding the appropriate boundedness condition.

3. Compact kryptonite produces heat at a rate of H cal/s cm3. If a sphere(radius c) of this material transfers heat by convection to a surroundingmedium at temperature T, the temperature u(ρ) in the sphere satisfies theboundary value problem

1

ρ2

d

dρ

(ρ2 du

dρ

)= −H

κ, 0 < ρ < c,

−κdu

dρ(c) = h

(u(c) − T

).

Supply the proper boundedness condition and solve. What is the tempera-ture at the center of the sphere?

4. (Critical radius) The neutron flux u in a sphere of uranium obeys the dif-ferential equation

λ

3

1

ρ2

d

dρ

(ρ2 du

dρ

)+ (k − 1)Au = 0

in the range 0 < ρ < a, where λ is the effective distance traveled by a neu-tron between collisions, A is called the absorption cross section, and k isthe number of neutrons produced by a collision during fission. In addition,the neutron flux at the boundary of the sphere is 0. Make the substitutionu = v/ρ and 3(k − 1)A/λ = µ2, and determine the differential equationsatisfied by v(ρ). See Section 0.1, Exercise 19.

5. Solve the equation found in Exercise 4 and then find u(ρ) that satisfiesthe boundary value problem (with boundedness condition) stated in Ex-ercise 4. For what radius a is the solution not identically 0?

0.5 Green’s Functions 43

6. Inside a nuclear fuel rod, heat is constantly produced by nuclear reaction.A typical rod is about 3 m long and about 1 cm in diameter, so temperaturevariation along the length is much less than along a radius. Thus, we treatthe temperature in such a rod as a function of the radial variable alone. Findthis temperature u(r), which is the solution of the boundary value problem

1

r

d

dr

(r

du

dr

)= − g

κ, 0 < r < a,

u(a) = T0.

7. For the problem of Exercise 6, find the temperature at the center of therod, u(0), using these values for the parameters: a = 0.5 cm, the powerdensity g = 418 W/cm3 = 100 cal/s cm3, conductivity κ = 0.01 cal/s cm ◦C,and the surface temperature T0 = 325◦C.

8. A model for microwave heating of food uses this equation for the tempera-ture u(x) in a large solid object:

d2u

dx2= −Ae−x/L, 0 < x.

Here, A is a constant representing the strength of the radiation and prop-erties of the object, and L is a characteristic length, known as penetrationdepth, that depends on frequency of the radiation and properties of the ob-ject. (Typically, L is about 12 cm in frozen raw beef or 2 cm thawed.) Showthat the boundary condition u′(0) = 0 is incompatible with the conditionthat u(x) be bounded as x goes to infinity. [See C.J. Coleman, The mi-crowave heating of frozen substances, Applied Math. Modeling, 14 (1990):439–443.]

9. Solve the differential equation in Exercise 8 subject to the conditions

u(0) = T0, u(x) bounded.

0.5 Green’s FunctionsThe most important features of the solution of the boundary value problem,1

d2u

dx2+ k(x)

du

dx+ p(x)u = f (x), l < x < r, (1)

αu(l) − α′u′(l) = 0, (2)

βu(r) + β ′u′(r) = 0, (3)

1The primes on the constants α′, β ′ are not to indicate differentiation, of course, but toshow that they are coefficients of derivatives.


can be developed by using the variation-of-parameters solution of the differ-ential equation (1), as presented in Section 2. To begin, we need to have twoindependent solutions of the homogeneous equation

d2u

dx2+ k(x)

du

dx+ p(x)u = 0, l < x < r. (4)

Let us designate these two solutions as u1(x) and u2(x). It will simplify algebralater if we require that u1 satisfy the boundary condition at x = l and u2 thecondition at x = r;

αu1(l) − α′u′1(l) = 0, (5)

βu2(r) + β ′u′2(r) = 0. (6)

According to Theorem 3 of Section 2, the general solution of the differentialequation (1) can be written as

u(x) = c1u1(x) + c2u2(x) +∫ x

l

(u1(z)u2(x) − u2(z)u1(x)

) f (z)

W(z)dz. (7)

Recall that in the denominator of the integrand, we have the Wronskian of u1

and u2,

W(z) =∣∣∣∣∣u1(z) u2(z)

u′1(z) u′

2(z)

∣∣∣∣∣ , (8)

which is nonzero because u1 and u2 are independent. We will need to knowthe following derivative of the function in Eq. (7):

du

dx= c1u′

1(x) + c2u′2(x) +

∫ x

l

(u1(z)u′

2(x) − u2(z)u′1(x)

) f (z)

W(z)dz.

(See Leibniz’s rule in the Appendix.)Now let us apply the boundary condition, Eq. (2), to the general solution

u(x). First, at x = l we have

αu(l) − α′u′(l) = c1

(αu1(l) − α′u′

1(l)) + c2

(αu2(l) − α′u′

2(l)) = 0. (9)

Note that the integrals in u and u′ are both 0 at x = l. Because of the boundarycondition (5) imposed on u1, Eq. (9) reduces to

c2

(αu2(l) − α′u′

2(l)) = 0, (10)

and we conclude that c2 = 0.


Second, the boundary condition at x = r becomes

βu(r) + β ′u′(r) = c1

(βu1(r) + β ′u′(r)

) +∫ r

l

[u1(z)

(βu2(r) + β ′u′

2(r))

− u2(z)(βu1(r) + β ′u′

1(r))] f (z)

W(z)dz = 0. (11)

Now, the boundary condition (6) on u2 at x = r eliminates one term of theintegrand, leaving

c1

(βu1(r) + β ′u′

1(r)) −

∫ r

lu2(z)

(βu1(r) + β ′u′

1(r)) f (z)

W(z)dz = 0. (12)

The common factor of βu1(r)+β ′u′1(r) can be canceled from both terms, and

we then find

c1 =∫ r

lu2(z)

f (z)

W(z)dz. (13)

Now we have found c1 and c2 so that u(x) in Eq. (7) satisfies both boundaryconditions. If we use the values of c1 and c2 as found, we have

u(x) = u1(x)

∫ r

lu2(z)

f (z)

W(z)dz

+∫ x

l

(u1(z)u2(x) − u2(z)u1(x)

) f (z)

W(z)dz. (14)

The solution becomes more compact if we break the interval of integration atx in the first integral, making it∫ r

lu2(z)

f (z)

W(z)dz =

∫ x

lu2(z)

f (z)

W(z)dz +

∫ r

xu2(z)

f (z)

W(z)dz. (15)

When the integrals on the range l to x are combined, there is some cancella-tion, and our solution becomes

u(x) =∫ x

lu1(z)u2(x)

f (z)

W(z)dz +

∫ r

xu1(x)u2(z)

f (z)

W(z)dz. (16)

Finally, these two integrals can be combined into one. We first define theGreen’s function for the problem (1), (2), (3) as

G(x, z) =

u1(z)u2(x)

W(z), l < z ≤ x,

u1(x)u2(z)

W(z), x ≤ z < r.

(17)


Then the formula given in Eq. (16) for u simplifies to

u(x) =∫ r

lG(x, z)f (z)dz. (18)

Example.Solve the problem that follows by constructing the Green’s function.

d2u

dx2− u = −1, 0 < x < 1,

u(0) = 0, u(1) = 0.

First, we must find two independent solutions of the homogeneous differentialequation u′′ − u = 0 that satisfy the boundary conditions as required. Thegeneral solution of the homogeneous differential equation is

u(x) = c1 cosh(x) + c2 sinh(x).

As u1(x) is required to satisfy the condition at the left, u1(0) = 0, we takec1 = 0, c2 = 1 and conclude u1(x) = sinh(x). The second solution is to sat-isfy u2(1) = 0. We may take

u2(x) = sinh(1) cosh(x) − cosh(1) sinh(x) = sinh(1 − x).

The Wronskian of the two solutions is

W(x) =∣∣∣∣ sinh(x) sinh(1 − x)cosh(x) − cosh(1 − x)

∣∣∣∣ = − sinh(1).

Now, by Eq. (17), the Green’s function for this problem is

G(x, z) =

sinh(z) sinh(1 − x)

− sinh(1), 0 < z ≤ x,

sinh(x) sinh(1 − z)

− sinh(1), x ≤ z < 1.

Furthermore, since f (x) = −1, the solution, by Eq. (18), is the integral

u(x) =∫ 1

0−G(x, z)dz.

To actually carry out the integration, we must break the interval of integrationat x, thus reverting in effect to Eq. (16). The result:


u(x) =∫ x

0

sinh(z) sinh(1 − x)

sinh(1)dz +

∫ 1

x

sinh(x) sinh(1 − z)

sinh(1)dz

= sinh(1 − x)

sinh(1)cosh(z)

∣∣∣∣x

0

+ sinh(x)

sinh(1)

(− cosh(1 − z))∣∣∣∣

1

x

= sinh(1 − x)

sinh(1)

(cosh(x) − 1

) + sinh(x)

sinh(1)

(cosh(1 − x) − 1

)

= sinh(1 − x) cosh(x) + sinh(x) cosh(1 − x)

sinh(1)− sinh(1 − x) + sinh(x)

sinh(1)

= 1 − sinh(1 − x) + sinh(x)

sinh(1).

This, finally, is easily seen to be the correct solution. In this instance, thereare much quicker ways to arrive at the same result. The advantage of theGreen’s function is that it shows how the solution of the problem dependson the inhomogeneity f (x). It is an efficient way to obtain the solution insome cases. �

Now let us look back over the calculations and see if there is some place theymight fail. Aside from the possibility that the coefficients k(x) or p(x) in thedifferential equation might not be continuous, it seems that division by 0 isthe only possibility of failure. Quantities canceled or divided by were

W(x) =∣∣∣∣∣u1(x) u2(x)

u′1(x) u′

2(x)

∣∣∣∣∣ ,αu2(l) − α′u′

2(l),

βu1(r) + β ′u′1(r)

in Eqs. (7), (10), and (12), respectively. It can be shown that all three of theseare 0 if any one of them is 0, and, in that case, u1(x) and u2(x) are proportional.We summarize in a theorem.

Theorem. Let k(x), p(x), and f (x) be continuous, l ≤ x ≤ r. The boundary valueproblem

d2u

dx2+ k(x)

du

dx+ p(x)u = f (x), l < x < r,

αu(l) − α′u′(l) = 0, (i)

βu(r) + β ′u′(r) = 0, (ii)


has one and only one solution, unless there is a nontrivial solution of

d2u

dx2+ k(x)

du

dx+ p(x)u = 0, l < x < r,

that satisfies (i) and (ii).When a unique solution exists, it is given by Eqs. (17) and (18). �

Example.The boundary value problem

d2u

dx2+ u = −1, 0 < x < π,

u(0) = 0, u(π) = 0,

does not have a unique solution, according to the theorem, because u(x) =sin(x) is a nontrivial solution of the problem

d2u

dx2+ u = 0, 0 < x < π,

u(0) = 0, u(π) = 0.

Indeed, if we try to follow through the construction, we find that u1(x) =sin(x) and also u2(x) = sin(x) (or a multiple thereof), and so all three quanti-ties in Eq. (19) are 0.

On the other hand, suppose we try to obtain a solution by the usual method.The general solution of the differential equation is

u(x) = −1 + c1 cos(x) + c2 sin(x).

However, application of the boundary conditions leads to the contradictoryrequirements

−1 + c1 = 0 and −1 − c1 = 0.

Thus, in this case, there simply is no solution to the problem stated. �

If the differential equation (1) has a singular point at x = l or x = r (orboth), a Green’s function may still be constructed. The boundary condition (2)or (3) would be replaced by a boundedness condition, which would also applyto u1 or u2 as the case may be.

Example.Construct Green’s function for the problem

1

x

d

dx

(x

du

dx

)= f (x), 0 < x < 1,

u(0) bounded, u(1) = 0.


The general solution of the corresponding homogeneous equation is u(x) =c1 + c2 ln(x). Thus, we would choose

u1(x) = 1, u2(x) = ln(x)

so that u1(x) is bounded at x = 0 and u2(x) is 0 at x = 1. The Green’s functionis thus

G(x, z) ={

z ln(x), 0 < z ≤ x,

z ln(z), x ≤ z < 1.

A similar procedure is followed if the interval l < x < r is infinite in length. �

E X E R C I S E S

In Exercises 1–8, find the Green’s function for the problem stated.

1.d2u

dx2= f (x), 0 < x < a,

u(0) = 0, u(a) = 0.

2.d2u

dx2= f (x), 0 < x < a,

u(0) = 0,du

dx(a) = 0.

3.d2u

dx2− γ 2u = f (x), 0 < x < a,

du

dx(0) = 0, u(a) = 0.

4.1

r

d

dr

(r

du

dr

)= f (r), 0 ≤ r < c,

u(c) = 0, u(r) bounded at r = 0.

5.1

ρ2

d

dρ

(ρ2 du

dρ

)= f (ρ), 0 ≤ ρ < c,

u(c) = 0, u(ρ) bounded at ρ = 0.

6.d2u

dx2+ 1

x

du

dx− 1

4x2u = f (x), 0 ≤ x < a,

u(a) = 0, u(x) bounded at x = 0.

7.d2u

dx2− γ 2u = f (x), 0 < x,

u(0) = 0, u(x) bounded as x → ∞.


8.d2u

dx2− γ 2u = f (x), −∞ < x < ∞,

u(x) bounded as x → ±∞.

9. Use the Green’s function of Exercise 5 to solve the problem

1

ρ2

d

dρ

(ρ2 du

dρ

)= 1, 0 ≤ ρ < c,

u(c) = 0,

and compare with the solution found by integrating the equation directly.

10. Use the Green’s function of Exercise 8 to solve the problem

d2u

dx2− γ 2u = −γ 2, −∞ < x < ∞,

u(x) bounded as x → ±∞,

and compare with the result found directly.

11. Use the Green’s function of Exercise 1 to solve the problem stated there, if

f (x) ={

0, 0 < x < a/2,

1, a/2 < x < a.

12. In confirmation of the theorem, show that the homogeneous problem ahas a nontrivial solution; problem b has no solution (existence fails); andproblem c has infinitely many solutions (uniqueness fails).

a. u′′ + u = 0, u(0) = 0, u(π) = 0,

b. u′′ + u = −1, u(0) = 0, u(π) = 0,

c. u′′ + u = π − 2x, u(0) = 0, u(π) = 0.

13. Considering z to be a parameter (l < z < r), define the function v(x) =G(x, z) with G as in Eq. (17). Show that v has these four properties, whichare sometimes used to define the Green’s function.

(i) v satisfies the boundary conditions, Eqs. (2) and (3), at x = l and r.

(ii) v is continuous, l < x < r. (The point x = z needs to be checked.)

(iii) v′ is discontinuous at x = z, and

limh→0+

(v′(z + h) − v′(z − h)

) = 1.

(iv) v satisfies the differential equation v′′ + k(x)v′ + p(x)v = 0 for l <

x < z and z < x < r.


14. Show that the boundary value problem

d2u

dx2+ λ2u = f (x), 0 < x < a,

u(0) = 0, u(a) = 0,

will have no solution or infinitely many solutions if λ is an eigenvalue of

d2u

dx2+ λ2u = 0,

u(0) = 0, u(a) = 0.

Chapter ReviewSee the CD for review questions.

Miscellaneous ExercisesIn Exercises 1–15, solve the given boundary value problem, supplying bound-edness conditions where necessary.

1.d2u

dx2− γ 2u = 0, 0 < x < a,

u(0) = T0, u(a) = T1.

2.d2u

dx2− r = 0, 0 < x < a (r is constant),

u(0) = T0,du

dx(a) = 0.

3.d2u

dx2= 0, 0 < x < a,

u(0) = T0,du

dx(a) = 0.

4.d2u

dx2− γ 2u = 0, 0 < x < a,

du

dx(0) = 0, u(a) = T1.

5.1

r

d

dr

(r

du

dr

)= −p, 0 < r < a,

u(a) = 0.


6.1

r

d

dr

(r

du

dr

)= 0, a < r < b,

u(a) = T0, u(b) = T1.

7.1

ρ2

d

dρ

(ρ2 du

dρ

)= −H, 0 < ρ < a,

u(a) = T0.

8.1

r

d

dr

(r

du

dr

)= 0, a < r < ∞,

u(a) = T.

9.d2u

dx2− γ 2(u − T) = 0, 0 < x < a,

du

dx(0) = 0, u(a) = T1.

10.d2u

dx2− γ 2u = 0, 0 < x < ∞,

u(0) = T.

11.d2u

dx2= γ 2(u − T0), 0 < x < ∞,

u(0) = T.

12.d

dx

(x3 du

dx

)= −k, a < x < b (k is constant),

u(a) = 0, u(b) = 0 (Note: 0 < a.)

13. In this problem, h is the groundwater level between two trenches inwhich water is held at constant levels. Solve for h(x). Note that the equa-tion is nonlinear.

d

dx

(h

dh

dx

)+ e = 0, 0 < x < a,

h(0) = h0, h(a) = h1.

14. Solve for u(x).

d4u

dx4= w, 0 < x < a (w is constant),

u(0) = 0, u(a) = 0,d2u

dx2(0) = 0,

d2u

dx2(a) = 0.


15. Solve for u(x). Note the interval.

d4u

dx4+ k

EIu = w, 0 < x < ∞ (w is constant),

u(0) = 0,d2u

dx2(0) = 0.

16. Show that any two of the four functions sinh(λx), sinh(λ(a − x)),cosh(λx), cosh(λ(a − x)) are independent solutions of the differentialequation

φ′′ − λ2φ = 0.

17. In this problem, u is the temperature in a wall composed of two sub-stances. Find u(x).

d2u

dx2= 0, 0 < x < αa and αa < x < a,

u(0) = T0, u(a) = T1,

κ1du

dx(αa−) = κ2

du

dx(αa+),

u(αa−) = u(αa+).

The last two conditions say that the heat flow rate and the temperatureare both continuous across the interface at x = αa.

18. Find the general solution of the differential equation

1

x2

d

dx

(x2 du

dx

)+ ku = 0

for the cases k = λ2 and k = −p2. (Hint: Let u(x) = v(x)/x and find theequation that v(x) satisfies.)

19. Find the solution of the boundary value problem

ex d

dx

(ex du

dx

)= −1, 0 < x < a,

u(0) = 0, u(a) = 0.

20. Solve the boundary value problem

1

r

d

dr

(r

du

dr

)= −rk, 0 < r < a,

u(0) bounded and u(a) = 0.


21. Solve the differential equation

d2u

dx2= p2u, 0 < x < a,

subject to the following sets of boundary conditions.

a. u(0) = 0, u(a) = 1;

b. u(0) = 1, u(a) = 0;

c. u′(0) = 0, u(a) = 1;

d. u(0) = 1, u′(a) = 0;

e. u′(0) = 1, u′(a) = 0;

f. u′(0) = 0, u′(a) = 1.

22. Solve the integro-differential boundary value problem

d2u

dx2= γ 2

(u −

∫ 1

0u(x)dx

), 0 < x < 1,

du

dx(0) = 0, u(1) = T.

Hint: Look for a solution in the form

u(x) = A cosh(γ x) + B sinh(γ x) + C.

23. Use a variation of parameters to find a second independent solution ofthe following differential equation. One solution is given in parentheses.

d2u

dx2− 2x

1 − x2

du

dx+ 2

1 − x2u = 0 (u = x).

24. By applying the method of variation of parameters, derive this formulafor a particular solution of the differential equation

d2u

dx2− γ 2u = f (x),

u(x) =∫ x

0f (x′)

sinhγ (x − x′)γ

dx′.

25. The absolute temperature u(x) in a cooling fin that radiates heat to amedium at absolute temperature T obeys the differential equation u′′ =γ 2(u4 − T4). Solve the special version in the boundary value problem


that follows, which can be done in closed form.

d2u

dx2= γ 2u4, 0 < x,

u(0) = U, limx→∞ u(x) = 0.

26. A uniform, straight shaft exhibits violent behavior at certain frequen-cies of rotation. Let the x-axis between 0 and a represent the undeflectedcenterline of the shaft, and let u(x) be the displacement of the actual cen-terline of the shaft measured from the x-axis. Centrifugal force providesa transverse loading on the shaft when u is not identically equal to zero.The equation for the displacement is

d4u

dx4− ω2w

EIgu = 0, 0 < x < a,

where w is the weight per unit length of the shaft, g is the accelerationof gravity, E is Young’s modulus, I is the second moment of the cross-sectional area of the shaft, and ω is the angular velocity. If the shaft isheld in narrow bearings at the ends, these can be interpreted as simplesupports, leading to boundary conditions

u(0) = 0, u′′(0) = 0, u(a) = 0, u′′(a) = 0.

Find a formula for those values of angular velocity (critical values orwhirling speeds) that permit the existence of nonzero solutions to thisboundary value problem.

27. Find the lowest critical value for the angular velocity of a steel shaft withthese specifications: diameter 1.5 in.; length 48 in.; w = 0.5 lb/in.; E =30 × 106 lb/in.2, I = 0.5 in.4.

28. Sulphur dioxide (SO2) is a common air pollutant that reacts with waterto form sulphuric acid. If the water is airborne, the result is acid rain; ifthe water is in snow, the result is acid runoff when the snow melts. Use ananalysis similar to that of Section 3 to obtain a boundary value problemfor the concentration u(x) (in units of mass per unit volume) of sulphurdioxide in the air included in a layer of snow. Introduce q(x), the flowrate of sulphur dioxide (in units of mass per unit time per unit of cross-sectional area.) There are two important physical facts: (1) Diffusion isgoverned by Fick’s law (similar to Fourier’s law)

q(x) = −Ddu

dx,


where D is the diffusion constant; and (2) when the sulphur dioxide reactswith water, it “disappears” at a rate proportional to its concentration, sayku(x) (in units of mass per unit time per unit volume).

29. The sulphur dioxide concentration in the air in a deep layer of snowsatisfies this boundary value problem in equilibrium conditions:

d2u

dx2− a2u = 0, 0 < x,

u(0) = C0.

Here, C0 is the concentration in freely circulating air. Add an appropriateboundedness condition and solve for u(x).

30. In “Mechanical properties of thin films from the load deflection of longclamped plates” [V. Ziebart et al., J. of Microelectromechanical Systems, 7(1998): 320–327] this boundary value problem is studied:

d4w

dx4− γ 2 d2w

dx2= P, −1

2< x <

1

2,

w

(±1

2

)= 0,

dw

dx

(±1

2

)= 0.

The variables are w deflection, x distance measured across the short di-mension; and the parameters are P pressure beneath the plate and γ 2

effective stress, all dimensionless. Find the general solution of the differ-ential equation.

31. (Continuation) Solve the boundary value problem in Exercise 30.

32. (Continuation) The parameter γ 2 is related to stress, which is related todeflection. It must satisfy the equation

γ 2 = S0 +∫ 1/2

0

(dw

dx

)2

dx.

Use your solution to find a single explicit equation that γ satisfies.

33. (Continuation) If S0 is negative in the equation of Exercise 32, γ 2 mightbe negative, say, γ 2 = −λ2. If this is the case, is there a value of λ forwhich the solution breaks down?

34. Suppose that u(t) is a function, not identically 0, for which

u′′

u= constant > 0.


Show that this relation is a differential equation and solve it. (Call theconstant p2.) Prove that exactly one of the following three possibilitiesholds:

(i) u(t) = 0 for one value of t and u′(t) is never 0;

(ii) u′(t) = 0 for one value of t and u(t) is never 0;

(iii) neither u(t) nor u′(t) is ever 0.

Fourier Series andIntegrals C H A P T E R

1

1.1 Periodic Functions and Fourier Series

A function f is said to be periodic with period p > 0 if: (1) f (x) has been de-fined for all x; and (2) f (x + p) = f (x) for all x. The familiar functions sin(x)and cos(x) are simple examples of periodic functions with period 2π , and thefunctions sin(2πx/p) and cos(2πx/p) are periodic with period p.

A periodic function has many periods, for if f (x) = f (x + p) then also

f (x) = f (x + p) = f (x + 2p) = · · · = f (x + np),

where n is any integer. Thus sin(x) has periods 2π,4π, . . . ,2nπ, . . . . The pe-riod of a periodic function is generally taken to be positive, but the periodicitycondition holds for negative as well as positive changes in the argument. Thatis to say, f (x − p) = f (x) for all x, since f (x) = f (x − p + p) = f (x − p). Also,

f (x) = f (x − p) = f (x − 2p) = · · · = f (x − np).

The definition of periodic says essentially that functional values repeatthemselves. This implies that the graph of a periodic function can be drawnfor all x by making a template of the graph on any interval of length p andthen copying the graph from the template up and down the x-axis (see Fig. 1).

Many of the functions that occur in engineering and physics are periodicin space or time — for example, acoustic waves — and in order to understandthem better it is often desirable to represent them in terms of the very simpleperiodic functions 1, sin(x), cos(x), sin(2x), cos(2x), and so forth. All of these

59

60 Chapter 1 Fourier Series and Integrals

Figure 1 A periodic function of period p.

functions have the common period 2π , although each has other periods aswell.

If f is periodic with period 2π , then we attempt to represent f in the formof an infinite series

f (x) = a0 +∞∑

n=1

(an cos(nx) + bn sin(nx)

). (1)

Each term of the series has period 2π , so if the sum of the series exists, it willbe a function of period 2π . There are two questions to be answered: (a) Whatvalues must a0, an, bn have? (b) If the appropriate values are assigned to thecoefficients, does the series actually represent the given function f (x)?

On the face of it, the first question is tremendously difficult, for Eq. (1) rep-resents an equation in an infinite number of unknowns. But a reasonable an-swer can be found easily by using the orthogonality1 relations shown in Table 1.We may summarize those relations by saying: The definite integral (over theinterval −π to π) of the product of any two different functions from the seriesin Eq. (1) is zero.

The fundamental idea is that if the equality proposed in Eq. (1) is to be a realequality, then both sides must give the same result after the same operation.The orthogonality relations then suggest operations that simplify the right-hand side of Eq. (1). Namely, we multiply both sides of the proposed equationby one of the functions that appears there and integrate from −π to π . (Wemust assume that the integration of the series can be carried out term by term.This is sometimes difficult to justify, but we do it nonetheless.)

Multiplying both sides of Eq. (1) by the constant 1 (= cos(0x)) and inte-grating from −π to π , we find

∫ π

−π

f (x)dx =∫ π

−π

a0 dx +∞∑

n=1

∫ π

−π


)dx.

1The word orthogonality should not be thought of in the geometric sense.

Chapter 1 Fourier Series and Integrals 61

∫ π

−π

sin(nx)dx = 0

∫ π

−π

cos(nx)dx ={

0, n �= 02π, n = 0∫ π

−π

sin(nx) cos(mx)dx = 0

∫ π

−π

sin(nx) sin(mx)dx ={

0, n �= mπ, n = m∫ π

−π

cos(nx) cos(mx)dx ={

0, n �= mπ, n = m �= 0.

Table 1 Orthogonality relations

Each of the terms in the integrated series is zero, so the right-hand side of thisequation reduces to 2π · a0, giving

a0 = 1

2π

∫ π

−π

f (x)dx.

(In words, a0 is the mean value of f (x) over one period.)Now multiplying each side of Eq. (1) by sin(mx), where m is a fixed integer,

and integrating from −π to π , we find

∫ π

−π

f (x) sin(mx)dx =∫ π

−π

a0 sin(mx)dx +∞∑

n=1

∫ π

−π

an cos(nx) sin(mx)dx

+∞∑

n=1

∫ π

−π

bn sin(nx) sin(mx)dx.

All terms containing a0 or an disappear, according to the orthogonality rela-tions. Furthermore, of all those containing a bn, the only one that is not zero isthe one in which n = m. (Notice that n is a summation index and runs throughall the integers 1,2, . . . . We chose m to be a fixed integer, so n = m once.) Wenow have the formula

bm = 1

π

∫ π

−π

f (x) sin(mx)dx.

By multiplying both sides of Eq. (1) by cos(mx) (m is a fixed integer) andintegrating, we also find

am = 1

π

∫ π

−π

f (x) cos(mx)dx.


We can now summarize our results. In order for the proposed equality

f (x) = a0 +∞∑

n=1


)(2)

to hold, the a’s and b’s must be chosen according to the formulas

a0 = 1

2π

∫ π

−π

f (x)dx, (3)

an = 1

π

∫ π

−π

f (x) cos(nx)dx, (4)

bn = 1

π

∫ π

−π

f (x) sin(nx)dx. (5)

When the coefficients are chosen this way, the right-hand side of Eq. (1) iscalled the Fourier series of f . The a’s and b’s are called Fourier coefficients. Wehave not yet answered question (b) about equality, so we write

f (x) ∼ a0 +∞∑

n=1


)

to indicate that the Fourier series corresponds to f (x). See the CD for an ani-mated example.

Example.Suppose that f (x) is periodic with period 2π and is given by the formulaf (x) = x in the interval −π < x < π (see Fig. 2). According to our formulas,

a0 = 1

2π

∫ π

−π

f (x)dx = 1

2π

∫ π

−π

x dx = 0,

an = 1

π

∫ π

−π

f (x) cos(nx)dx = 1

π

∫ π

−π

x cos(nx)dx

= 1

π

[cos(nx)

n2+ x sin(nx)

n

]∣∣∣∣π

−π

= 0,

bn = 1

π

∫ π

−π

f (x) sin(nx)dx = 1

π

∫ π

−π

x sin(nx)dx

= 1

π

[sin(nx)

n2− x cos(nx)

n

]∣∣∣∣π

−π

= 1

π

(−2π) cos nπ

n= 2

n(−1)n+1.

Chapter 1 Fourier Series and Integrals 63

Figure 2 f (x) = x, −π < x < π , f periodic with period 2π .

Thus, for this function, we have

f (x) ∼∞∑

n=1

2(−1)n+1

nsin(nx)

∼ 2

(sin(x) − 1

2sin(2x) + 1

3sin(3x) − · · ·

). �

The Appendix contains some integration formulas that are convenient forfinding Fourier coefficients. It is also useful to know these special values ofsines and cosines that come up frequently in Fourier series.

sin(nπ) = 0, cos(nπ) = (−1)n, for n = 0,±1,±2, . . . ,

sin

((2n − 1)π

2

)= (−1)n+1, cos

((2n − 1)π

2

)= 0,

for n = 0,±1,±2, . . . .

Note that the second line involves only odd multiples of π/2. Even multiplesof π/2 are included in the first line.

E X E R C I S E S

1. Find the Fourier coefficients of the functions given in what follows. All aresupposed to be periodic with period 2π . Sketch the graph of the function.

a. f (x) = x, −π < x < π ;

b. f (x) = |x|, −π < x < π ;

c. f (x) ={0, −π < x < 0,

1, 0 < x < π ;

d. f (x) = |sin x|.2. Sketch for at least two periods the graphs of the functions defined by:

a. f (x) = x, −1 < x ≤ 1, f (x + 2) = f (x);


b. f (x) ={0, −1 < x ≤ 0,

x, 0 < x < 1,f (x + 2) = f (x);

c. f (x) ={0, −π < x ≤ 0,

1, 0 < x ≤ 2π ,f (x + 3π) = f (x);

d. f (x) ={0, −π < x ≤ 0,

sin x, 0 < x ≤ π ,f (x + 2π) = f (x).

3. Show that the constant function f (x) = 1 is periodic with every possibleperiod p > 0.

4. Carry out the details of deriving the equation for am.

5. Suppose f (x) has period p. Show that for any c, the following equationholds. Hint: Think of the integral as the net signed area.∫ c+p

cf (x)dx =

∫ p

0f (x)dx.

6. Suppose f (x), g(x) are periodic with a common period p. Show that af (x)+bg(x) and f (x) · g(x) also are periodic with period p (a, b are constants).

7. Find the Fourier series of each of the following periodic functions. Integra-tion is not necessary: Use trigonometric identities.

a. f (x) = cos2(x);

b. f (x) = sin(x − π/6);

c. f (x) = sin(x) cos(2x).

8. Verify that sin(πx/a) and cos(πx/a) are periodic with period 2a.

1.2 Arbitrary Period and Half-Range ExpansionsIn Section 1 we found a way to represent a periodic function of period2π with a Fourier series. It is not necessary to restrict ourselves to this pe-riod. In fact, we may broaden the idea of Fourier series to include func-tions of any period by a simple rescaling of the variables. Let us supposethat a function f is periodic with period 2a. (We use 2a in place of p forlater convenience.) Then we may relate f to a series of the functions 1,sin(πx/a), cos(πx/a), sin(2πx/a), cos(2πx/a), . . . , all having period 2a, inthe form

f (x) ∼ a0 +∞∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

).

The coefficients of this Fourier series may be determined either by scaling fromthe formulas of Section 1 or through the concept of orthogonality. In either

1.2 Arbitrary Period and Half-Range Expansions 65

case, the coefficients are

a0 = 1

2a

∫ a

−af (x)dx, an = 1

a

∫ a

−af (x) cos

(nπx

a

)dx,

bn = 1

a

∫ a

−af (x) sin

(nπx

a

)dx.

(1)

Example.Find the Fourier series of f (x) = |sin(πx)|.Solution: This function is periodic with period 1, so a = p/2 = 1/2. To do theintegrals for the Fourier coefficients, we need to get rid of the absolute valuesigns:

f (x) ={

sin(πx), 0 < x < 1,

− sin(πx), −1 < x < 0.

Then, it is easy to calculate

a0 = 1

1

∫ 1/2

−1/2

∣∣sin(πx)∣∣ dx =

∫ 0

−1/2− sin(πx)dx +

∫ 1/2

0sin(πx)dx

= cos(πx)

π

∣∣∣∣0

−1/2

−cos(πx)

π

∣∣∣∣1/2

0

= 1

π− −1

π= 2

π.

(Recall that cos(±π/2) = 0.) The other coefficients are found similarly:

an = 2

1

∫ 1/2

−1/2

∣∣sin(πx)∣∣cos(2nπx)dx

= 2

[∫ 0

−1/2− sin(πx) cos(2nπx)dx +

∫ 1/2

0sin(πx) cos(2nπx)dx

]

= − 4

π· 1

4n2 − 1.

And bn is found to be 0 for all n. Consequently, the Fourier series of the functionis

∣∣sin(πx)∣∣ ∼ 2

π− 4

π

∞∑n=1

1

4n2 − 1cos(2nπx).

We will see later that |sin(πx)| is equal to its series. �

It is often necessary to use a Fourier series to represent a function that hasbeen defined only in a finite interval. We can justify such a representation bymaking the given function part of a periodic function. If the given function f isdefined on the interval −a < x < a, we may construct f , the periodic extension


of period 2a, by using the following definitions:

f (x) = f (x), −a < x < a,

f (x) = f (x + 2a), −3a < x < −a,

f (x) = f (x − 2a), a < x < 3a

and so on, up and down the x-axis. Notice that the argument of f on the right-hand side always falls in the interval −a < x < a, where f was originally given.Graphically, this kind of extension amounts to making a template of the graphof f on −a < x < a and then copying from the template in abutting intervalsof length 2a.

For the extended function with period 2a, the formulas for the Fourier co-efficients become

a0 = 1

2a

∫ a

−af (x)dx,

an = 1

a

∫ a

−af (x) cos

(nπx

a

)dx,

bn = 1

a

∫ a

−af (x) sin

(nπx

a

)dx.

(2)

If we are concerned with f (x) only in the interval −a < x < a where it wasoriginally given, the process of periodic extension is strictly formal, becausethe formulas for the coefficients involve f only on the original interval. Thus,we may write

f (x) ∼ a0 +∞∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

), −a < x < a.

The inequality for x draws attention to the fact that f was defined only on theinterval −a to a.

Example.Suppose f (x) = x in the interval −1 < x < 1. The graph of its periodic exten-sion (with period 2) is seen in Fig. 3, and the Fourier coefficients are

a0 = 0, an = 0,

bn =∫ 1

−1x sin(nπx)dx = −2 cos(nπ)

nπ= 2

π

(−1)n+1

n. �

The sine and cosine functions that appear in a Fourier series have somespecial symmetry properties that are useful in evaluating the coefficients. Thegraph of the cosine function is symmetric about the vertical axis, and that ofthe sine is antisymmetric. We formalize these properties with a definition.


Figure 3 f (x) = x, −1 < x < 1, f periodic with period 2.

DefinitionA function g(x) is even if g(−x) = g(x); h(x) is odd if h(−x) = −h(x). Notethat a function must be defined on a symmetric interval, say −c < x < c(where c might be ∞), in order to qualify as even or odd.

An even function is often said to be symmetric about the vertical axis, and anodd function is said to be symmetric in the origin. Many familiar functions areeither even or odd. For example, sin(kx), x, x3, and any other odd power of xare all odd functions defined on the interval −∞ < x < ∞. Similarly, cos(kx),|x|, 1 (= x0), x2, and any other even power of x are even functions over thesame interval. Most functions are neither even nor odd, but any function thatis defined on a symmetric interval can be written as a sum of an even and anodd function:

f (x) = 1

2

(f (x) + f (−x)

) + 1

2

(f (x) − f (−x)

).

It is easy to show that the first term is an even function and the second is odd.Even and odd functions preserve their symmetries in some algebraic opera-

tions, as summarized here:

even + even = even, odd + odd = odd,

even × even = even, odd × odd = even, odd × even = odd.

We are also concerned with definite integrals of even and odd functions oversymmetric intervals. The symmetry properties lead to important simplifica-tions in our calculations.

Theorem 1. Let g(x) be an even function defined in a symmetric interval−a < x < a. Then ∫ a

−ag(x)dx = 2

∫ a

0g(x)dx.


Let h(x) be an odd function defined in a symmetric interval −a < x < a. Then∫ a

−ah(x)dx = 0. �

Suppose now that g is an even function in the interval −a < x < a. Since thesine function is odd and the product g(x) sin(nπx/a) is odd,

bn = 1

a

∫ a

−ag(x) sin

(nπx

a

)dx = 0.

That is, all the sine coefficients are zero. Also, since the cosine is even, so isg(x) cos(nπx/a), and then

an = 1

a

∫ a

−ag(x) cos

(nπx

a

)dx = 2

a

∫ a

0g(x) cos

(nπx

a

)dx.

Thus the cosine coefficients can be computed from an integral over the intervalfrom 0 to a.

Parallel results hold for odd functions: the cosine coefficients are all zeroand the sine coefficients can be simplified. We summarize the results.

Theorem 2. If g(x) is even on the interval −a < x < a (g(−x) = g(x)), then

g(x) ∼ a0 +∞∑

n=1

an cos

(nπx

a

), −a < x < a,

where

a0 = 1

a

∫ a

0g(x)dx, an = 2

a

∫ a

0g(x) cos

(nπx

a

)dx.

If h(x) is odd on the interval −a < x < a (h(−x) = −h(x)), then

h(x) ∼∞∑

n=1

bn sin

(nπx

a

), −a < x < a,

where

bn = 2

a

∫ a

0h(x) sin

(nπx

a

)dx. �

Very frequently, a function given in an interval 0 < x < a must be repre-sented in the form of a Fourier series. There are infinitely many ways of do-ing this, but two ways are especially simple and useful: extending the givenfunction to one defined on a symmetric interval −a < x < a by making theextended function either odd or even.


(a) (b)

(c) (d)

Figure 4 A function is given in the interval 0 < x < a (heavy curve). The figureshows: (a) the odd extension; (b) the even extension; (c) the odd periodic exten-sion; and (d) the even periodic extension.

DefinitionLet f (x) be given for 0 < x < a. The odd extension of f is defined by

fo(x) ={

f (x), 0 < x < a,−f (−x), −a < x < 0.

The even extension of f is defined by

fe(x) ={

f (x), 0 < x < a,f (−x), −a < x < 0.

Notice that if −a < x < 0, then 0 < −x < a, so the functional values on theright are known from the given functions.

Graphically, the even extension is made by reflecting the graph in the verticalaxis. The odd extension is made by reflecting first in the vertical axis and thenin the horizontal axis (see Fig. 4).

Now the Fourier series of either extension may be calculated from the for-mulas in Theorem 2. Since fe is even and fo is odd, we have

fe(x) ∼ a0 +∞∑

n=1

an cos

(nπx

a

), −a < x < a,

fo(x) ∼∞∑

n=1

bn sin

(nπx

a

), −a < x < a.


If the series on the right converge, they actually represent periodic functionswith period 2a. The cosine series would represent the even periodic extensionof f — the periodic extension of fe; and the sine series would represent the oddperiodic extension of f .

When the problem at hand is to represent the function f (x) in the interval0 < x < a, where it was originally given, we may use either the Fourier sineseries or the cosine series because both fe and fo coincide with f in the interval.

Thus we may summarize by saying: If f (x) is given for 0 < x < a, then

f (x) ∼ a0 +∞∑

n=1

an cos

(nπx

a

), 0 < x < a,

a0 = 1

a

∫ a

0f (x)dx, an = 2

a

∫ a

0f (x) cos

(nπx

a

)dx

and

f (x) ∼∞∑

n=1

bn sin

(nπx

a

), 0 < x < a,

bn = 2

a

∫ a

0f (x) sin

(nπx

a

)dx.

These two representations are called half-range expansions, and the series arecalled the Fourier cosine and Fourier sine series of f , respectively. We shallneed these, more than any other kind of Fourier series, in the applications wemake later in this book.

Example.Let us suppose that the function f has the formula

f (x) = x, 0 < x < 1.

Then the odd periodic extension of f is as shown in Fig. 5, and the Fourier sinecoefficients of f are

bn = 2

∫ 1

0x sin(nπx)dx = − 2

nπcos(nπ).

The even periodic extension of f is shown in Fig. 6. The Fourier cosine co-efficients are

a0 =∫ 1

0x dx = 1

2,

an = 2

∫ 1

0x cos(nπx)dx = − 2

n2π2

(1 − cos(nπ)

). �


Figure 5 Odd periodic extension (period 2) of f (x) = x, 0 < x < 1.

Figure 6 Even periodic extension (period 2) of f (x) = x, 0 < x < 1.

The following six correspondences (we will later show them to be equalities)follow from the ideas of this section. Note that the inequalities showing theapplicable range of x are crucial.

∞∑n=1

−2 cos(nπ)

nπsin(nπx) ∼

f (x) = x, 0 < x < 1,

fo(x) = x, −1 < x < 1,

fo(x), −∞ < x < ∞,

1

2−

∞∑n=1

2(1 − cos(nπ))

n2π2cos(nπx) ∼

f (x) = x, 0 < x < 1,

fe(x) = |x|, −1 < x < 1,

fe(x), −∞ < x < ∞.

E X E R C I S E S

1. Find the Fourier series of each of the following functions. Sketch the graphof the periodic extension of f for at least two periods.

a. f (x) = |x|, −1 < x < 1;

b. f (x) ={−1, −2 < x < 0,

1, 0 < x < 2;

c. f (x) = x2, − 12 < x < 1

2 .


2. Show that the functions cos(nπx/a) and sin(nπx/a) satisfy orthogonalityrelations similar to those given in Section 1.

3. Suppose a Fourier series is needed for a function defined in the interval0 < x < 2a. Show how to construct a periodic extension with period 2a,and give formulas for the Fourier coefficients that use only integrals from0 to 2a. (Hint: See Exercise 5, Section 1.)

4. Show that the formula

ex = cosh(x) + sinh(x)

gives the decomposition of the function ex into a sum of an even and anodd function.

5. Identify each of the following as being even, odd, or neither. Sketch on asymmetric interval.

a. f (x) = x;

c. f (x) = |cos(x)|;e. f (x) = x cos(x);

b. f (x) = |x|;d. f (x) = arc sin(x);

f. f (x) = x+cos(x+1).

6. If f (x) is given in the interval 0 < x < a, what other ways are there toextend it to a function on −a < x < a?

7. Find the Fourier series of these functions.

a. f (x) = x, −1 < x < 1;

b. f (x) = 1, −2 < x < 2;

c. f (x) ={

x, − 12 < x < 1

2 ,

1 − x, 12 < x < 3

2 .

8. Is it true that if all the sine coefficients of a function f defined on −a< x < a are zero, then f is even?

9. We know that if f (x) is odd on the interval −a < x < a, its Fourier se-ries is composed only of sines. What additional symmetry condition on fwill make the sine coefficients with even indices be zero? Give an exam-ple.

10. Sketch both the even and odd extensions of these functions.

a. f (x) = 1, 0 < x < a;

c. f (x) = sin(x), 0 < x < 1;

b. f (x) = x, 0 < x < a;

d. f (x) = sin(x), 0 < x < π .

11. Find the Fourier sine series and cosine series for the functions given inExercise 10. Sketch the even and odd periodic extensions for several peri-ods.

1.3 Convergence of Fourier Series 73

12. Prove the orthogonality relations∫ a

0sin

(nπx

a

)sin

(mπx

a

)dx =

{0, n �= m,a/2, n = m,

∫ a

0cos

(mπx

a

)cos

(nπx

a

)dx =

{0, n �= m,a/2, n = m �= 0,a, n = m = 0.

13. If f (x) is continuous on the interval 0 < x < a, is its even periodic ex-tension continuous? What about the odd periodic extension? Check espe-cially at x = 0 and ±a.

14. Justify Theorem 1 by considering the integral as a sum of signed areas. SeeFig. 4 for typical even and odd functions.

15. Justify or prove these statements.

a. If h(x) is an odd function, then |h(x)| is an even function.

b. If f (x) is defined for all positive x, then f (|x|) is an even func-tion.

c. If f (x) is defined for all x and g(x) is any even function, then f (g(x)) iseven.

d. If h(x) is an odd function, g(x) is even, and g(x) is defined for all x,then g(h(x)) is an even function.

1.3 Convergence of Fourier SeriesNow we are ready to take up the second question of Section 1: Does the Fourierseries of a function actually represent that function? The word represent hasmany interpretations, but for most practical purposes we really want to knowthe answer to this question: If a value of x is chosen, the numbers cos(nπx/a)and sin(nπx/a) are computed for each n and inserted into the Fourier seriesof f , and the sum of the series is calculated, is that sum equal to the functionalvalue f (x)?

In this section we shall state, without proof, some theorems that answer thequestion (a proof of the convergence theorem is given in Section 7). But firstwe need a few definitions about limits and continuity.

The ordinary limit limx→x0 f (x) can be rewritten as limh→0 f (x0 + h). Hereh may approach zero in any manner. But if h is required to be positive only, weget what is called the right-hand limit of f at x0, defined by

f (x0+) = limh→0+

f (x0 + h) = limh→0h>0

f (x0 + h).


(a) (b) (c)

Figure 7 Three functions with different kinds of discontinuities at x = 1.(a) f (x) = (x − x2)/(1 − x) has a removable discontinuity. (b) f (x) = x for0 < x < 1 and f (x) = x − 1 for 1 < x; this function has a jump discontinuity.(c) f (x) = − ln(|1 − x|) has a “bad” discontinuity.

The left-hand limit is defined similarly:

f (x0−) = limh→0−

f (x0 + h) = limh→0h<0

f (x0 + h) = limh→0+

f (x0 − h).

Note that f (x0+) and f (x0−) need not be values of the function f .If both left- and right-hand limits exist and are equal, the ordinary limit

exists and is equal to the one-handed limits. It is quite possible that the left-and right-handed limits exist but are different. This happens, for instance, atx = 0 for the function

f (x) ={1, 0 < x < π ,

−1, −π < x < 0.

In this case, the left-hand limit at x0 = 0 is −1, whereas the right-hand limitis +1. A discontinuity at which the one-handed limits exist but do not agree iscalled a jump discontinuity.

It is also possible that at some point both limits exist and agree but that thefunction is not defined at that point or its value is not equal to the limit. Insuch a case, a function is said to have a removable discontinuity. If the value ofthe function at the troublesome point is redefined to be equal to the limit, thefunction will become continuous. For example, the function f (x) = sin(x)/xhas a removable discontinuity at x = 0. The discontinuity is eliminated by re-defining f (x) = sin(x)/x (x �= 0), f (0) = 1. Removable discontinuities are sosimple that we may assume they have been removed from any function underdiscussion.

Other discontinuities are more serious. They occur if one or both of theone-handed limits fail to exist. Each of the functions sin(1/x), e1/x, 1/x has adiscontinuity at x = 0 that is neither removable nor a jump (see Fig. 7). Table 2summarizes continuity behavior at a point.


Name CriterionContinuity f (x0+) = f (x0−) = f (x0)

Removable discontinuity f (x0+) = f (x0−) �= f (x0)

Jump discontinuity f (x0+) �= f (x0−)

“Bad” discontinuity f (x0+) or f (x0−) or both fail to exist

Table 2 Types of continuity behavior at x0

Figure 8 Typical sectionally continuous function made up of four continuous“sections.”

We shall say that a function is sectionally continuous (also called piecewisecontinuous) on an interval a < x < b if it is bounded and continuous, ex-cept possibly for a finite number of jumps and removable discontinuities. (SeeFig. 8.) A function is sectionally continuous (without qualification) if it is sec-tionally continuous on every interval of finite length. For instance, if a periodicfunction is sectionally continuous on any interval whose length is one periodor more, then it is sectionally continuous.

Examples.

1. The square wave, defined by

f (x) ={1, 0 < x < a,

−1, −a < x < 0,f (x + 2a) = f (x),

is sectionally continuous. There are jump discontinuities at x = 0, ±a,±2a, etc.

2. The function f (x) = 1/x cannot be sectionally continuous on any intervalthat contains 0 or even has 0 as an endpoint, because the function is notbounded at x = 0.

3. If f (x) = x, −1 < x < 1, then f is continuous on that interval. Its periodicextension (see Fig. 3) is sectionally continuous but not continuous. �


The examples clarify a couple of facts about the meaning of sectional con-tinuity. Most important is that a sectionally continuous function must not“blow up” at any point — even an endpoint — of an interval. Note also that afunction need not be defined at every point in order to qualify as sectionallycontinuous. No value was given for the square-wave function at x = 0, ±a,but the function remains sectionally continuous, no matter what values areassigned for these points.

A function is sectionally smooth (also, piecewise smooth) in an interval a <

x < b if: f is sectionally continuous; f ′(x) exists, except perhaps at a finitenumber of points; and f ′(x) is sectionally continuous. The graph of a section-ally smooth function then has a finite number of removable discontinuities,jumps, and corners. (The derivative will not exist at these points.) Betweenthese points, the graph will be continuous, with a continuous derivative. Novertical tangents are allowed, for these indicate that the derivative is infinite.

Examples.

1. f (x) = |x|1/2 is continuous but not sectionally smooth in any interval thatcontains 0, because |f ′(x)| → ∞ as x → 0.

2. The square wave is sectionally smooth but not continuous. �

Most of the functions useful in mathematical modeling are sectionallysmooth. Fortunately we can also give a positive statement about the Fourierseries of such functions.

Theorem. If f (x) is sectionally smooth and periodic with period 2a, then at eachpoint x the Fourier series corresponding to f converges, and its sum is

a0 +∞∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

)= f (x+) + f (x−)

2. �

See an animated example on the CD.This theorem gives an answer to the question at the beginning of the section.

Recall that a sectionally smooth function has only a finite number of jumpsand no bad discontinuities in every finite interval. Hence,

f (x−) = f (x+) = 1

2

(f (x+) + f (x−)

) = f (x),

except perhaps at a finite number of points on any finite interval. For this rea-son, if f satisfies the hypotheses of the theorem, we write f equal to its Fourierseries, even though the equality may fail at jumps.

In constructing the periodic extension of a function, we never defined thevalues of f (x) at the endpoints. Since the Fourier coefficients are given by in-tegrals, the value assigned to f (x) at one point cannot influence them; in that


sense, the value of f at x = +a is unimportant. But because of the averagingfeatures of the Fourier series, it is reasonable to define

f (a) = f (−a) = 1

2

(f (a−) + f (−a+)

).

That is, the value of f at the endpoints is the average of the one-handed limitsat the endpoints, each limit taken from the interior. For instance, if f (x) =1 + x, 0 < x < 1, and f (x) = 0, −1 < x < 0, then f (±1) should be taken tobe 1, and f (0) should be 1

2 .

Examples.

1. The square-wave function

f (x) ={1, 0 < x < 1,

−1, −1 < x < 0

is sectionally smooth; therefore the corresponding Fourier series con-verges to {

1, for 0 < x < 1,−1, for −1 < x < 0,0, for x = 0,1,−1

and is periodic with period 2.

2. For the function f (x) = |x|1/2, −π < x < π , f (x + 2π) = f (x), the pre-ceding theorem does not guarantee convergence of the Fourier series atany point, even though the function is continuous. Nevertheless, the se-ries does converge at any point x! This shows that the conditions in thetheorem are perhaps too strong. (But they are useful.) �

E X E R C I S E S

1. For each function given, if it is not sectionally smooth on the interval, ex-plain why not. Sketch.

a. f (x) = |x| − |1 − x|, −1 < x < 2;

b. f (x) = √|x|, −1 < x < 1;

c. f (x) = ln(2 cos(x/2)

), −π < x < π ;

d. f (x) = tan(x), 0 < x < π/2;

e. f (x) = tan(x), 0 < x < π .

2. Check each function described in what follows to see whether it is section-ally smooth. If it is, state the value to which its Fourier series converges ateach point x in the interval and at the endpoints. Sketch.


a. f (x) = |x| + x, −1 < x < 1;

b. f (x) = x cos(x), −π

2< x <

π

2;

c. f (x) = x cos(x), −1 < x < 1;

d. f (x) ={0, 1 < x < 3,

1, −1 < x < 1,x, −3 < x < −1.

3. To what value does the Fourier series of f converge if f is a continuous,sectionally smooth, periodic function? Give an example.

4. State convergence theorems for the Fourier sine and cosine series that arisefrom half-range expansions.

5. A function is given on the interval 0 < x < 2 by the formula

f (x) ={ x, 0 < x < 1,

1 − x, 1 < x < 2.

a. Sketch the odd periodic extension f0(x) for −4 < x < 4.

b. Explain why f0(x) is sectionally smooth.

c. Determine the value that the sine series of f converges to at these points:x = 1, x = 2, x = 9.6, x = −3.8.

6. For the same function given in Exercise 5, answer the same questions forfe(x), the even periodic extension of f and its cosine series.

7. The series∞∑

n=1

(−1)n

n2cos(nx)

converges to a function f (x) whose formula on the interval −π < x < π is

f (x) = A + Bx + Cx2.

Determine A,B, and C.

8. The series∞∑

n=1

1

n3sin(nx)

converges to a continuous periodic function. On the interval 0 < x < 2π ,this function coincides with a polynomial p(x) of degree 3. Find the polyno-mial. Hint: Determine points x on the interval 0 < x < 2π where p(x) = 0.Use this information to get a form for p(x).

1.4 Uniform Convergence 79

9. The function f (x) is periodic with period 2. Its graph for −1 < x < 1 is asemicircle with radius 1 centered at the origin.

a. Find the equation of f (x) for −1 < x < 1.

b. Determine the value of the coefficient a0 in its Fourier series. (This is theonly cosine coefficient that can be found in closed form.)

c. Is f (x) sectionally smooth?

d. What does the theorem tell us about the convergence of the Fourier se-ries of f (x)?

1.4 Uniform ConvergenceThe theorem of the preceding section treats convergence at individual pointsof an interval. A stronger kind of convergence is uniform convergence in aninterval. Let

SN(x) = a0 +N∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

)

be the partial sum of the Fourier series of a function f . The maximum devia-tion between the graphs of SN(x) and f (x) is

δN = max∣∣ f (x) − SN(x)

∣∣, −a ≤ x ≤ a,

where the maximum2 is taken over all x in the interval, including the end-points. If the maximum deviation tends to zero as N increases, we say that theseries converges uniformly in the interval −a ≤ x ≤ a.

Roughly speaking, if a Fourier series converges uniformly, then the sum of afinite number N of terms gives a good approximation — to within ±δN — ofthe value of f (x) at any and every point of the interval. Furthermore, by takinga large enough N , one can make the error as small as necessary.

There are two important facts about uniform convergence. If a Fourier seriesconverges uniformly in a period interval, then (1) it must converge to a con-tinuous function, and (2) it must converge to the (continuous) function thatgenerates the series. Thus, a function that has a nonremovable discontinuitycannot have a uniformly convergent Fourier series. (And not all continuousfunctions have uniformly convergent Fourier series.)

Figure 9 presents graphs of some partial sums of a square-wave function. Itis easy to see that for every N there are points near x = 0 and x = ±π where

2If f is not continuous, the maximum must be replaced by the supremum, or least upperbound.


Figure 9 Partial sums of the square-wave function. Convergence is not uniform.


Figure 10 Partial sums of a sawtooth function. Convergence is uniform.


| f (x) − SN(x)| is nearly equal to 1, so convergence is not uniform. (Inciden-tally, the graphs in Fig. 9 also show the partial sums of f (x) overshooting theirmark near x = 0. This feature of Fourier series is called Gibbs’ phenomenonand always occurs near a jump.) On the other hand, Fig. 10 shows graphs of a“sawtooth” function and the partial sums of its Fourier series. The maximumdeviation always occurs at x = 0, and the convergence is uniform.

One of the ways of proving uniform convergence is by examining the coef-ficients.

Theorem 1. If the series∑∞

n=1(|an| + |bn|) converges, then the Fourier series

a0 +∞∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

)

converges uniformly in the interval −a ≤ x ≤ a and, in fact, on the whole interval−∞ < x < ∞. �

Example.For the function

f (x) = |x|, −π < x < π,

the Fourier coefficients are

a0 = π

2, an = 2

π

cos(nπ) − 1

n2, bn = 0.

Since the series∑∞

n=1 1/n2 converges, the series of absolute values of the coeffi-cients converges, and so the Fourier series converges uniformly on the interval−π ≤ x ≤ π to |x|. The Fourier series converges uniformly to the periodicextension of f (x) on the whole real line (see Fig. 10). �

Another way of proving uniform convergence of a Fourier series is by exam-ining the function f that generates it.

Theorem 2. If f is periodic and continuous and has a sectionally continuousderivative, then the Fourier series corresponding to f converges uniformly to f (x)on the entire real axis. �

While this theorem is stated for a periodic function, it may be adapted to afunction f (x) given on the interval −a < x < a. If the periodic extension of fsatisfies the conditions of the theorem, then the Fourier series of f convergesuniformly on the interval −a ≤ x ≤ a.

Example.Consider the function

f (x) = x, −1 < x < 1.


Although f (x) is continuous and has a continuous derivative in the interval−1 < x < 1, the periodic extension of f is not continuous. The Fourier seriescannot converge uniformly in any interval containing 1 or −1 because the pe-riodic extension of f has jumps there, but uniform convergence must producea continuous function.

On the other hand, the function f (x) = |sin(x)|, periodic with period 2π , iscontinuous and has a sectionally continuous derivative. Therefore, its Fourierseries converges uniformly to f (x) everywhere. �

Here is a restatement of Theorem 2 for a function given on the interval−a < x < a. The condition at the endpoints replaces the condition of conti-nuity of the periodic extension of f .

Theorem 3. If f (x) is given on −a < x < a, if f is continuous and bounded andhas a sectionally continuous derivative, and if f (−a+) = f (a−), then the Fourierseries of f converges uniformly to f on the interval −a ≤ x ≤ a. (The series con-verges to f (a−) = f (−a+) at x = ±a.) �

If an odd periodic function is to be continuous, it must have value 0 at x = 0and at the endpoints of the symmetric period-interval. Thus, the odd periodicextension of a function given in 0 < x < a may have jump discontinuities eventhough it is continuous where originally given. The even periodic extensioncauses no such difficulty, however.

Theorem 4. If f (x) is given on 0 < x < a, if f is continuous and bounded and hasa sectionally continuous derivative, and if f (0+) = f (a−) = 0, then the Fouriersine series of f converges uniformly to f in the interval 0 ≤ x ≤ a. (The seriesconverges to 0 at x = 0 and x = a.) �

Theorem 5. If f (x) is given on 0 < x < a and if f is continuous and boundedand has a sectionally continuous derivative, then the Fourier cosine series of fconverges uniformly to f in the interval 0 ≤ x ≤ a. (The series converges to f (0+)

at x = 0 and to f (a−) at x = a.) �

E X E R C I S E S

1. Determine whether the Fourier series of the following functions convergeuniformly or not. Sketch each function.

a. f (x) = ex, −1 < x < 1;

b. f (x) = sinh(x), −π < x < π ;

c. f (x) = sin(x), −π < x < π ;


d. f (x) = sin(x) + |sin(x)|, −π < x < π ;

e. f (x) = x + |x|, −π < x < π ;

f. f (x) = x(x2 − 1), −1 < x < 1;

g. f (x) = 1 + 2x − 2x3, −1 < x < 1.

2. The Fourier series of the function

f (x) = sin(x)

x, −π < x < π,

converges at every point. To what value does the series converge at x = 0?at x = π ? The convergence is uniform. Why?

3. Determine whether the sine and cosine series of the following functionsconverge uniformly. Sketch.

a. f (x) = sinh(x), 0 < x < π ;

b. f (x) = sin(x), 0 < x < π ;

c. f (x) = sin(πx), 0 < x < 12 ;

d. f (x) = 1/(1 + x), 0 < x < 1;

e. f (x) = 1/(1 + x2), 0 < x < 2.

4. If an and bn tend to zero as n tends to infinity, show that the series

a0 +∞∑

n=1

e−αn(an cos(nx) + bn sin(nx)

)

converges uniformly (α > 0).

5. For each of the following coefficients, use Theorem 1 to decide whetherconvergence of the associate Fourier series is uniform.

a. an = sin2(nπ/2)

n2π2, bn = 0;

b. an = 0, bn = 1 − cos(nπ)

nπ;

c. a1 = 0, an = 2(1 + cos(nπ))

n2 − 1(n ≥ 2), bn = 0;

d. an = 0, bn = 1

cosh(nπ/2).

1.5 Operations on Fourier Series 85

1.5 Operations on Fourier Series

In the course of this book we shall have to perform certain operations onFourier series. The purpose of this section is to find conditions under whichthey are legitimate. Two things must be noted, however. First, the theoremsstated here are not the best possible: There are theorems with weaker hypothe-ses and the same conclusions. Second, in applying mathematics, we often carryout operations formally, legitimate or not. The results must then be checkedfor correctness.

Throughout this section we shall state results about functions and Fourierseries with period 2π , for typographic convenience. The results remain truewhen the period is 2a instead. For functions defined only on a finite interval,the periodic extension must fulfill the hypotheses. We shall refer to a functionf (x) with the series shown:

f (x) ∼ a0 +∞∑

n=1

an cos(nx) + bn sin(nx). (1)

Theorem 1. The Fourier series of the function cf (x) has coefficients ca0, can, andcbn (c is constant). �

This theorem is a simple consequence of the fact that a constant passesthrough an integral. The fact that the integral of a sum is the sum of the inte-grals leads to the following.

Theorem 2. The Fourier coefficients of the sum f (x) + g(x) are the sums of thecorresponding coefficients of f (x) and g(x). �

These two theorems are so natural that the reader has probably used themalready without thinking about it. The theorems that follow are much moredifficult to prove, but they are extremely important.

Theorem 3. If f (x) is periodic and sectionally continuous, then the Fourier seriesof f may be integrated term by term:

∫ b

af (x)dx =

∫ b

aa0 dx +

∞∑n=1

∫ b

a


)dx. (2)

�

Theorem 4. If f (x) is periodic and sectionally continuous and if g(x) is sectionallycontinuous for a ≤ x ≤ b, then

86 Chapter 1 Fourier Series and Integrals∫ b

af (x)g(x)dx =

∫ b

aa0g(x)dx

+∞∑

n=1

∫ b

a


)g(x)dx. (3)

�

In Theorems 3 and 4, the function f (x) is only required to be sectionallycontinuous. It is not necessary that the Fourier series of f (x) converge at all.Nevertheless, the theorems guarantee that the series on the right converges andequals the integral on the left in Eqs. (2) and (4).

One important application of Theorem 4 was the derivation of the formulasfor the Fourier coefficients in Section 1. An application of Theorems 3 and 4is given in what follows.

Example.The periodic function g(x) whose formula in the interval 0 < x < 2π is

g(x) = x, 0 < x < 2π

has the Fourier series

g(x) ∼ π − 2∞∑

n=1

sin(nx)

n.

By applying Theorems 1 and 2, we find that the function f (x) defined by f (x) =[π − g(x)]/2 has the series

f (x) ∼∞∑

n=1

sin(nx)

n.

This manipulation would be simple algebra if the correspondence ∼ were anequality.

The function f (x) satisfies the hypotheses of Theorem 3. Thus we may inte-grate the preceding series from 0 to b to obtain

∫ b

0f (x)dx =

∞∑n=1

1 − cos(nb)

n2.

Theorem 3 guarantees that this equality holds for any b. In the interval from 0to 2π we have the formula f (x) = (π − x)/2. Hence

∫ b

0f (x)dx = πb

2− b2

4=

∞∑n=1

1 − cos(nb)

n2, 0 ≤ b ≤ 2π.


Now, replacing b by x, we have

x(2π − x)

4=

∞∑n=1

1

n2−

∞∑n=1

cos(nx)

n2, 0 ≤ x ≤ 2π. (4)

Outside the indicated interval, the periodic extension of the function on theleft equals the series on the right.

It is worthwhile to mention that the series on the right of Eq. (4) is theFourier series of the function on the left. That is to say,

1

2π

∫ 2π

0

x(2π − x)

4dx =

∞∑n=1

1

n2, (5)

1

π

∫ 2π

0

x(2π − x)

4cos(nx)dx = −1

n2, (6)

1

π

∫ 2π

0

x(2π − x)

4sin(nx)dx = 0. (7)

Equations (6) and (7) can be verified directly, of course, but Theorem 4, to-gether with the orthogonality relations of Section 1, also guarantees them. Inaddition, Eq. (5) gives us a way to evaluate the series on the right. �

Although the uniqueness property stated in the following theorem is so verynatural that we tend to assume it is true without checking, it really is a conse-quence of Theorem 4.

Theorem 5. If f (x) is periodic and sectionally continuous, its Fourier series isunique. �

That is to say, only one series can correspond to f (x). We often make use ofuniqueness in this way: If two Fourier series are equal (or correspond to thesame function), then the coefficients of like terms must match.

The last operation to be discussed is differentiation, one that plays a princi-pal role in applications.

Theorem 6. If f (x) is periodic, continuous, and sectionally smooth, then the dif-ferentiated Fourier series of f (x) converges to f ′(x) at every point x where f ′′(x)exists:

f ′(x) =∞∑

n=1

(−nan sin(nx) + nbn cos(nx)). (8)

�

The hypotheses on f (x) itself imply (see Section 4) that the Fourier series off (x) converges uniformly. If f (x) (or its periodic extension) fails to be contin-


uous, it is certain that the differentiated series of f (x) will fail to converge, atsome points at least.

Example.Let f be the function that is periodic with period 2π and has the formula

f (x) = |x|, −π < x < π.

This function is indeed continuous and sectionally smooth and is equal to itsFourier series,

f (x) = π

2− 4

π

(cos(x) + cos(3x)

9+ cos(5x)

25+ · · ·

).

According to Theorem 5, the differentiated series

4

π

(sin(x) + sin(3x)

3+ sin(5x)

5+ · · ·

)

converges to f ′(x) at any point x where f ′′(x) exists. Now, the derivative of thesawtooth function f (x) (see Fig. 10) is the square-wave function

f ′(x) ={1, 0 < x < π ,

−1, −π < x < 0(9)

(see Fig. 9). Moreover, we know that the foregoing sine series is the Fourierseries of the square wave f ′(x) and that it converges to the values given byEq. (9), except at the points where f ′(x) has a jump. These are precisely thepoints where f ′′(x) does not exist. �

Later on, it will frequently happen that we know a function only through itsFourier series. Thus, it will be important to obtain properties of the functionby examining its coefficients, as the next theorem does.

Theorem 7. If f is periodic, with Fourier coefficients an, bn, and if the series

∞∑n=1

(∣∣nkan

∣∣ + ∣∣nkbn

∣∣)

converges for some integer k ≥ 1, then f has continuous derivatives f ′, . . . , f (k)

whose Fourier series are differentiated series of f . �

Example.Consider the function defined by the series

f (x) =∞∑

n=1

e−nα cos(nx),


in which α is a positive parameter. For this function we have a0 = 0, an = e−nα ,bn = 0. By the integral test, the series

∑nke−nα converges for any k. Therefore

f has derivatives of all orders. The Fourier series of f ′ and f ′′ are

f ′(x) =∞∑

n=1

−ne−nα sin(nx),

f ′′(x) =∞∑

n=1

−n2e−nα cos(nx). �

E X E R C I S E S

1. Evaluate the sum of the series∑∞

n=1 1/n2 by performing the integrationindicated in Eq. (5).

2. Sketch the graphs of the periodic extension of the function

f (x) = π − x

2, 0 < x < 2π,

and of its derivative f ′(x) and of

F(x) =∫ x

0f (t)dt.

3. Suppose that a function has the formula f (x) = x, 0 < x < π . What is itsderivative? Can the Fourier sine series of f be differentiated term by term?What about the cosine series?

4. Verify Eqs. (6) and (7) by integration.

5. Suppose that a function f (x) is continuous and sectionally smooth in theinterval 0 < x < a. What additional conditions must f (x) satisfy in orderto guarantee that its sine series can be differentiated term by term? thecosine series?

6. Is the derivative of a periodic function periodic? Is the integral of a peri-odic function periodic?

7. It is known that the equality

ln

(∣∣∣∣2 cos

(x

2

)∣∣∣∣)

=∞∑

n=1

(−1)n+1

ncos(nx)

is valid except when x is an odd multiple of π . Can the Fourier series bedifferentiated term by term?


8. Use the series that follows, together with integration or differentiation, tofind a Fourier series for the function p(x) = x(π − x), 0 < x < π .

x = 2∞∑

n=1

(−1)n+1

nsin(nx), 0 < x < π.

9. Let f (x) be an odd, periodic, sectionally smooth function with Fouriersine coefficients b1,b2, . . . . Show that the function defined by

u(x, t) =∞∑

n=1

bne−n2t sin(nx), t ≥ 0,

has the following properties:

a.∂2u

∂x2=

∞∑n=1

−n2bne−n2t sin(nx), t > 0;

b. u(0, t) = 0, u(π, t) = 0, t > 0;

c. u(x,0) = 1

2

(f (x+) + f (x−)

).

10. Let f be as in Exercise 9, but define u(x, y) by

u(x, y) =∞∑

n=1

bne−ny sin(nx), y > 0.

Show that u(x, y) has these properties:

a.∂2u

∂x2=

∞∑n=1

−n2bne−ny sin(nx), y > 0;

b. u(0, y) = 0, u(π, y) = 0, y > 0;

c. u(x,0) = 1

2

(f (x+) + f (x−)

).

1.6 Mean Error and Convergence in MeanWhile we can study the behavior of infinite series, we must almost always usefinite series in practice. Fortunately, Fourier series have some properties thatmake them very useful in this setting. Before going on to these properties, weshall develop a useful formula.

Suppose f is a function defined in the interval −a < x < a, for which∫ a

−a

(f (x)

)2dx

1.6 Mean Error and Convergence in Mean 91

is a finite number. Let

f (x) ∼ a0 +∞∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

)

and let g(x) have a finite Fourier series

g(x) = A0 +N∑1

An cos

(nπx

a

)+ Bn sin

(nπx

a

).

Then we may perform the following operations:

∫ a

−af (x)g(x)dx =

∫ a

−af (x)

[A0 +

N∑1

An cos

(nπx

a

)+ Bn sin

(nπx

a

)]dx

= A0

∫ a

−af (x)dx +

N∑1

An

∫ a

−af (x) cos

(nπx

a

)dx

+N∑1

Bn

∫ a

−af (x) sin

(nπx

a

)dx.

We recognize the integrals as multiples of the Fourier coefficients of f andrewrite

1

a

∫ a

−af (x)g(x)dx = 2a0A0 +

N∑1

(anAn + bnBn). (1)

Now suppose we wish to approximate f (x) by a finite Fourier series. Thedifficulty here is deciding what “approximate” means. Of the many ways wecan measure approximation, the one that is easiest to use is the following:

EN =∫ a

−a

(f (x) − g(x)

)2dx. (2)

(Here g is the function with a Fourier series containing terms up to and in-cluding cos(Nπx/a).) Clearly, EN can never be negative, and if f and g are“close,” then EN will be small. Thus our problem is to choose the coefficientsof g so as to minimize EN . (We assume N fixed.)

To compute EN , we first expand the integrand:

EN =∫ a

−af 2(x)dx − 2

∫ a

−af (x)g(x)dx +

∫ a

−ag2(x)dx. (3)

The first integral has nothing to do with g; the other two integrals clearly de-pend on the choice of g and can be manipulated so as to minimize EN . We


already have an expression for the middle integral. The last one can be foundby replacing f with g in Eq. (1):

∫ a

−ag2(x)dx = a

[2A2

0 +N∑1

A2n + B2

n

]. (4)

Now we have a formula for EN in terms of the variables A0, An, Bn:

EN =∫ a

−af 2(x)dx − 2a

[2A0a0 +

N∑1

Anan + Bnbn

]

+a

[2A2

0 +N∑1

A2n + B2

n

]. (5)

The error EN takes its minimum value when all of the partial derivativeswith respect to the variables are zero. We must then solve the equations

∂EN

∂A0= −4aa0 + 4aA0 = 0,

∂EN

∂An= −2aan + 2aAn = 0,

∂EN

∂Bn= −2abn + 2aBn = 0.

These equations require that A0 = a0, An = an, Bn = bn. Thus g should bechosen to be the truncated Fourier series of f ,

g(x) = a0 +N∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

)

in order to minimize EN .Now that we know which choice of A’s and B’s minimizes EN , we can com-

pute that minimum value. After some algebra, we see that

min(EN) =∫ a

−af 2(x)dx − a

[2a2

0 +N∑1

a2n + b2

n

]. (6)

Even this minimum error must be greater than or equal to zero, and thus wehave the Bessel inequality

1

a

∫ a

−af 2(x)dx ≥ 2a2

0 +N∑1

a2n + b2

n. (7)

1.6 Mean Error and Convergence in Mean 93

This inequality is valid for any N and therefore is also valid in the limit as Ntends to infinity. The actual fact is that, in the limit, the inequality becomesParseval’s equality:

1

a

∫ a

−af 2(x)dx = 2a2

0 +∞∑1

a2n + b2

n. (8)

Another very important consequence of Bessel’s inequality is that the twoseries

∑a2

n and∑

b2n must converge if the left-hand side of Eqs. (7) and (8) is

finite. Thus, the numbers an and bn must tend to 0 as n tends to infinity.By comparing Eqs. (6) and (8), we get a different expression for the mini-

mum error:

min(EN) = a∞∑

N+1

a2n + b2

n.

This quantity decreases steadily to zero as N increases. Since min(EN) is, ac-cording to Eq. (2), a mean deviation between f and the truncated Fourier se-ries of f , we often say, “The Fourier series of f converges to f in the mean.”(Another kind of convergence!)

SummaryIf f (x) has been defined in the interval −a < x < a and if∫ a

−af 2(x)dx

is finite, then:

1. Among all finite series of the form

g(x) = A0 +N∑1

An cos

(nπx

a

)+ Bn sin

(nπx

a

)

the one that best approximates f in the sense of the error described byEq. (2) is the truncated Fourier series of f :

a0 +N∑1

an cos

(nπx

a

)+ bn sin

(nπx

a

).

2.1

a

∫ a

−af 2(x)dx = 2a2

0 +∞∑1

a2n + b2

n.


3. an = 1

a

∫ a

−af (x) cos

(nπx

a

)dx → 0 as n → ∞;

bn = 1

a

∫ a

−af (x) sin

(nπx

a

)dx → 0 as n → ∞.

4. The Fourier series of f converges to f in the sense of the mean.

Properties 2 and 3 are very useful for checking computed values of Fouriercoefficients.

E X E R C I S E S

1. Use properties of Fourier series to evaluate the definite integral

1

π

∫ π

−π

(ln

∣∣∣∣2 cos

(x

2

)∣∣∣∣)2

dx.

(Hint: See Section 10, Eq. (4), and Section 5, Eq. (5).)

2. Verify Parseval’s equality for these functions:

a. f (x) = x, −1 < x < 1;

b. f (x) = sin(x), −π < x < π .

3. What can be said about the behavior of the Fourier coefficients of the fol-lowing functions as n → ∞?

a. f (x) = |x|1/2, −1 < x < 1;

b. f (x) = |x|−1/2, −1 < x < 1.

4. How do we know that EN has a minimum and not a maximum?

5. If a function f defined on the interval −a < x < a has Fourier coefficients

an = 0, bn = 1√n,

what can you say about ∫ a

−af 2(x)dx?

6. Show that, as n → ∞, the Fourier sine coefficients of the function

f (x) = 1

x, −π < x < π,

tend to a nonzero constant. (Since this is an odd function, we can take thecosine coefficients to be zero, although strictly speaking they do not exist.)

1.7 Proof of Convergence 95

Use the fact that ∫ ∞

0

sin(t)

tdt = π

2.

1.7 Proof of ConvergenceIn this section we prove the Fourier convergence theorem stated in Section 3.Most of the proof requires nothing more than simple calculus, but there arethree technical points that we state here.

Lemma 1. For all N = 1,2, . . . ,

1

π

∫ π

−π

(1

2+

N∑n=1

cos(ny)

)dy = 1. �

Lemma 2. For all N = 1,2, . . . ,

1

2+

N∑n=1

cos(ny) = sin((N + 1

2 )y)

2 sin( 12 y)

. �

Lemma 3. If φ(y) is sectionally continuous, −π < y < π , then its Fourier coef-ficients tend to 0 with n:

limn→∞

1

π

∫ π

−π

φ(y) cos(ny)dy = 0,

limn→∞

1

π

∫ π

−π

φ(y) sin(ny)dy = 0. �

In Exercises 1 and 2 of this section, you are asked to verify Lemmas 1 and 2(also see Miscellaneous Exercise 17 at the end of this chapter). Lemma 3 wasproved in Section 6.

The theorem we are going to prove is restated here for easy reference. Period2π is used for typographic convenience; we have seen that any other periodcan be obtained by a simple change of variables.

Theorem. If f (x) is sectionally smooth and periodic with period 2π , then theFourier series corresponding to f converges at every x, and the sum of the series is

a0 +∞∑

n=1

an cos(nx) + bn sin(nx) = 1

2

(f (x+) + f (x−)

). (1)

�


Proof: Let the point x be chosen; it is to remain fixed. To begin with, weassume that f is continuous at x, so the sum of the series should be f (x).Another way to say this is that

limN→∞ SN(x) − f (x) = 0,

where SN is the partial sum of the Fourier series of f ,

SN(x) = a0 +N∑

n=1

an cos(nx) + bn sin(nx). (2)

Of course, the a’s and b’s are the Fourier coefficients of f ,

a0 = 1

2π

∫ π

−π

f (z)dz,

an = 1

π

∫ π

−π

f (z) cos(nz)dz,

bn = 1

2π

∫ π

−π

f (z) sin(nz)dz.

(3)

The integrals have z as their variable of integration, but that does not affecttheir value.

Part 1. Transformation of SN(x).In order to show a relationship between SN(x) and f , we replace the co-efficients in Eq. (2) by the integrals that define them and use elementaryalgebra on the results:

SN(x) = 1

2π

∫ π

−π

f (z)dx +N∑

n=1

[1

π

∫ π

−π

f (z) cos(nz)dz cos(nx)

+ 1

π

∫ π

−π

f (z) sin(nz)dz sin(nx)

](4)

= 1

2π

∫ π

−π

f (z)dx +N∑

n=1

[1

π

∫ π

−π

f (z) cos(nz) cos(nx)dz

+ 1

π

∫ π

−π

f (z) sin(nz) sin(nx)dz

](5)

= 1

2π

∫ π

−π

f (z)dx +N∑

n=1

[1

π

∫ π

−π

f (z)(cos(nz) cos(nx)

+ sin(nz) sin(nx))

dz

](6)


= 1

π

∫ π

−π

f (z)

(1

2+

N∑n=1

cos(nz) cos(nx) + sin(nz) sin(nx)

)dz

(7)

= 1

π

∫ π

−π

f (z)

(1

2+

N∑n=1

cos(n(z − x)

))dz. (8)

In this very compact formula for SN(x), we now change the variable of in-tegration from z to y = z − x:

SN(x) = 1

π

∫ π+x

−π+xf (x + y)

(1

2+

N∑n=1

cos(ny)

)dy. (9)

Note that both factors in the integrand are periodic with period 2π . Theinterval of integration can be any interval of length 2π with no change inthe result. (See Exercise 5 of Section 1.) Therefore,

SN(x) = 1

π

∫ π

−π

f (x + y)

(1

2+

N∑n=1

cos(ny)

)dy. (10)

Part 2. Expression for SN(x) − f (x).Since we must show that the difference SN(x) − f (x) goes to 0, we need tohave f (x) in a form compatible with that for SN(x). Recall that x is fixed(although arbitrary), so f (x) is to be thought of as a number. Lemma 1suggests the appropriate form,

f (x) = f (x) · 1

π

∫ π

−π

(1

2+

N∑n=1

cos(ny)

)dy

= 1

π

∫ π

−π

f (x)

(1

2+

N∑n=1

cos(ny)

)dy. (11)

Now, using Eq. (10) to represent SN(x), we have

SN(x) − f (x) = 1

π

∫ π

−π

(f (x + y) − f (x)

)(1

2+

N∑n=1

cos(ny)

)dy. (12)

Part 3. The limit.The next step is to use Lemma 2 to replace the sum in Eq. (12). The resultis

SN(x) − f (x) = 1

π

∫ π

−π

(f (x + y) − f (x)

) sin((N + 1

2 )y)

2 sin( 12 y)

dy. (13)


The addition formula for sines gives the equality

sin

((N + 1

2

)y

)= cos(Ny) sin

(1

2y

)+ sin(Ny) cos

(1

2y

).

Substituting it in Eq. (13) and using simple properties of integrals, we ob-tain

SN(x) − f (x) = 1

π

∫ π

−π

(f (x + y) − f (x)

)1

2cos(Ny)dy

+ 1

π

∫ π

−π

(f (x + y) − f (x)

) cos( 12 y)

2 sin( 12 y)

sin(Ny)dy. (14)

The first integral in Eq. (14) can be recognized as the Fourier cosine coeffi-cient of the function

ψ(y) = 1

2

(f (x + y) − f (x)

). (15)

Since f is a sectionally smooth function, so is ψ , and the first integral haslimit 0 as N increases, by Lemma 3.

The second integral in Eq. (14) can also be recognized, as the Fourier sinecoefficient of the function

φ(y) = f (x + y) − f (x)

2 sin( 12 y)

cos

(1

2y

). (16)

To proceed as before, we must show that φ(y) is at least sectionally contin-uous, −π ≤ y ≤ π . The only difficulty is to show that the apparent divisionby 0 at y = 0 does not cause φ(y) to have a bad discontinuity there.

First, if f is continuous and differentiable near x, then f (x + y) − f (x) iscontinuous and differentiable near y = 0. Then L’Hôpital’s rule gives

limy→0

f (x + y) − f (x)

2 sin( 12 y)

= limy→0

f ′(x + y)

cos( 12 y)

= f ′(x). (17)

Under these conditions, the function φ(y) of Eq. (16) has a removable dis-continuity at y = 0 and thus is sectionally continuous.

Second, if f is continuous at x but has a corner there, then f (x + y)− f (x)is continuous with a corner at y = 0. In this case, L’Hôpital’s rule applieswith the one-sided limits, which show

limy→0+

f (x + y) − f (x)

2 sin( 12 y)

= limy→0+

f ′(x + y)

cos( 12 y)

= f ′(x+), (18)

limy→0−

f (x + y) − f (x)

2 sin( 12 y)

= limy→0−

f ′(x + y)

cos( 12 y)

= f ′(x−). (19)


Under these conditions, the function φ(y) of Eq. (16) has a jump disconti-nuity at y = 0 and again is sectionally continuous.

In either case, we see that the second integral in Eq. (14) is the Fouriersine coefficient of a sectionally continuous function. By Lemma 3, then, ittoo has limit 0 as N increases, and the proof is complete for every x wheref is continuous.

Part 4. If f is not continuous at x.Now let us suppose that f has a jump discontinuity at x. In this case, wemust return to Part 2 and express the proposed sum of the series as

1

2

(f (x+) + f (x−)

) = 1

π

∫ π

0f (x+)

(1

2+

N∑n=1

cos(ny)

)dy

+ 1

π

∫ 0

−π

f (x−)

(1

2+

N∑n=1

cos(ny)

)dy. (20)

Here, we have used the evenness of the integrand in Lemma 1 to write

1

π

∫ π

0

(1

2+

N∑n=1

cos(ny)

)dy = 1

π

∫ 0

−π

(1

2+

N∑n=1

cos(ny)

)dy = 1

2. (21)

Next, we have a convenient way to write the quantity to be limited:

SN(x) − 1

2

(f (x+) + f (x−)

)

= 1

π

∫ π

0

(f (x + y) − f (x+)

)(1

2+

N∑n=1

cos(ny)

)dy

+ 1

π

∫ 0

−π

(f (x + y) − f (x−)

)(1

2+

N∑n=1

cos(ny)

)dy. (22)

The interval of integration for SN(x) as shown in Eq. (10) has been split inhalf to conform to the integrals in Eq. (20).

The last step is to show that each of the integrals in Eq. (22) approaches 0as N increases. Since the technique is the same as in Part 3, this is left as anexercise.

Let us emphasize that the crux of the proof is to show that the functionfrom Eq. (16),

φ(y) = f (x + y) − f (x)

2 sin( 12 y)

cos

(1

2y

)(23)

(or a similar function that arises from the integrands in Eq. (22)), does nothave a bad discontinuity at y = 0. �


E X E R C I S E S

1. Verify Lemma 2. Multiply through by 2 sin( 12 y). Use the identity

sin

(1

2y

)cos(ny) = 1

2

(sin

((n + 1

2

)y

)− sin

((n − 1

2

)y

)).

Note that most of the series then disappears. (To see this, write out theresult for N = 3.)

2. Verify Lemma 1 by integrating the sum term by term.

3. Let f (x) = f (x + 2π) and f (x) = |x| for −π < x < π . Note that f is con-tinuous and has a corner at x = 0. Sketch the function φ(y) as defined inEq. (16) if x = 0. Find φ(0+) and φ(0−).

4. Let f be the odd periodic extension of the function whose formula is π − xfor 0 < x < π . In this case, f has a jump discontinuity at x = 0. Takingx = 0, sketch the functions

φR(y) = f (x + y) − f (x+)

2 sin( 12 y)

cos

(1

2y

)(y > 0),

φL(y) = f (x + y) − f (x−)

2 sin( 12 y)

cos

(1

2y

)(y < 0).

(These functions appear if the integrands in Eq. (22) are developed as inPart 3 of the proof.)

5. Consider the function f that is periodic with period 2π and has the formulaf (x) = |x|3/4 for −π < x < π .

a. Show that f is continuous at x = 0 but is not sectionally smooth.

b. Show that the function φ(y) (from Eq. (16), with x = 0) is sectionallycontinuous, −π < x < π , except for a bad discontinuity at y = 0.

c. Show that the Fourier coefficients of φ(y) tend to 0 as n increases, de-spite the bad discontinuity.

1.8 Numerical Determination of Fourier CoefficientsThere are many functions whose Fourier coefficients cannot be determinedanalytically because the integrals involved are not known in terms of easilyevaluated functions. Also, it may happen that a function is not known explic-itly but that its value can be found at some points. In either case, if a Fourier

1.8 Numerical Determination of Fourier Coefficients 101

series is to be found for the function, some numerical technique must be em-ployed to approximate the integrals that give the Fourier coefficients. It turnsout that one of the crudest numerical integration techniques is the best.

Any periodic, sectionally smooth function can be reduced by the procedureillustrated in Fig. 11 to the sum of some functions f1(x) and f2(x), whose seriescan be found by integration, and another function that is continuous, periodic,and sectionally smooth. This last function’s Fourier coefficients will approach0 rapidly with n.

Suppose then that f (x) is continuous, sectionally smooth, and periodic withperiod 2a. We wish to find its Fourier coefficients numerically. For instance,

a0 = 1

2a

∫ a

−af (x)dx.

The integral is approximated using the trapezoidal rule. First, cut up the inter-val −a < x < a into r equal subintervals with endpoints x0, x1, . . . , xr where

xk = −a + kx, x = 2a

r.

Next, evaluate the sum

a0∼= 1

2a

(1

2f (x0) + f (x1) + · · · + f (xr−1) + 1

2f (xr)

)x. (1)

Since x0 = −a, xr = a, and f is periodic with period 2a, we have f (x0) = f (xr):The two terms with 1

2 multipliers can be combined. Thus, our approxima-tion is

a0∼= 1

2a

(f (x1) + f (x2) + · · · + f (xr)

) · 2a

r.

The occurrences of 2a cancel, and the computed value is just the average of thefunctional values.

We use a caret over the usual coefficient name to designate approximations.Other Fourier coefficients are approximated in a similar way.

SummaryLet f (x) be continuous, sectionally smooth and periodic with period 2a. Ap-proximate Fourier coefficients of f (x) are

a0 = 1

r

(f (x1) + · · · + f (xr)

), (2)

an = 2

r

(f (x1) cos

(nπx1

a

)+ · · · + f (xr) cos

(nπxr

a

)), (3)

bn = 2

r

(f (x1) sin

(nπx1

a

)+ · · · + f (xr) sin

(nπxr

a

)). (4)


Figure 11 Preparation of a function for numerical integration of Fourier co-efficients. (a) Graph of sectionally smooth function f (x) given on −a < x < a.(b) Graph of f1(x), which has jumps of the same magnitude and position as f (x).Coefficients can be found analytically. (c) Graph of f (x)− f1(x). This function hasno jumps in −a < x < a. (d) Graph of f2(x). The periodic extensions of f2(x) andof f (x)− f1(x) have jumps of the same magnitude at x = ±a, and so forth. The co-efficients of f2 can be found analytically. (e) Graph of f3(x) = f (x) − f1(x) − f2(x).The Fourier series of f3(x) converges uniformly (the coefficients tend to zerorapidly).


If r is odd, Eqs. (3) and (4) are valid for n = 1,2, . . . , (r −1)/2, giving a total ofr coefficients. If r is even, Eq. (4) gives br/2 = 0, and Eq. (3) has to be modified:

ar/2 = 1

r

(f (x1) cos

(r πx1

2a

)+ · · · + f (xr) cos

(r πxr

2a

)). (3′)

We again get r valid coefficients. �

The formulas in Eqs. (2)–(4) were derived for the case in which x0, x1, . . . , xr

are equally spaced points in the interval −a ≤ x ≤ a. However, they remainvalid for equally spaced points on the interval 0 ≤ x ≤ 2a. That is,

x0 = 0, x1 = 2a

r, x2 = 4a

r, . . . , xr = 2a. (5)

Note also that when f (x) is given in the interval 0 ≤ x ≤ a and the sine or co-sine coefficients are to be determined, the formulas may be derived from thosealready given here. Let the interval be divided into s equal subintervals withendpoints 0 = x0, x1, . . . , xs = a (in general, xi = ia/s). Then the approximateFourier cosine coefficients for f or its even extension are

a0 = 1

s

(1

2f (x0) + f (x1) + · · · + f (xs−1) + 1

2f (xs)

),

an = 2

s

(1

2f (x0) + f (x1) cos

(nπx1

a

)+ · · · + 1

2f (xs) cos

(nπxs

a

)),

n = 1, . . . , s − 1,

as = 1

s

(1

2f (x0) + f (x1) cos

(sπx1

a

)+ · · · + 1

2f (xs) cos

(sπxs

a

)). (6)

Similarly, the approximate Fourier sine coefficients for f or its odd extensionare

bn = 2

s

(f (x1) sin

(nπx1

a

)+ · · · + f (xs−1) sin

(nπxs−1

a

)),

n = 1,2, . . . , s. (7)

An important feature of the approximate Fourier coefficients is this: If

F(x) = a0 + a1 cos

(πx

a

)+ b1 sin

(πx

a

)+ · · ·

is a finite Fourier series using a total of r approximate coefficients calculatedfrom Eqs. (3) and (4), then F(x) actually interpolates the function f (x) atx1, x2, . . . , xr . That is,

F(xi) = f (xi), i = 1,2, . . . , r.


i xi cos xi cos 2xi cos 3xi sin(xi)/xi

0 0 1.0 1.0 1.0 1.01 π

60.86603 0.5 0 0.95493

2 π

30.5 −0.5 −1.0 0.82699

3 π

20 −1.0 0 0.63662

4 2π

3−0.5 −0.5 1.0 0.41350

5 5π

6−0.86603 0.5 0 0.19099

6 π −1.0 1.0 −1.0 0.0

Table 3 Numerical information

n an an Error0 0.58717 0.58949 0.002321 0.45611 0.45141 0.004702 −0.06130 −0.05640 0.004903 0.02884 0.02356 0.00528

Table 4 Approximate coefficientsof sin(x)/x

Thus the graph of F(x) cuts the graph of f (x) at the points xi, i = 1,2, . . . , r.

Example.Calculate the approximate Fourier coefficients of f (x) = sin(x)/x in −π <

x < π . Since f is even, it will have a cosine series. We simplify computation byusing the half-range formulas and making s even. We take s = 6, x0 = 0, x1 =π/6, . . . , x5 = 5π/6, x6 = π . The numerical information is given in Table 3.

The results of the calculation are given in Table 4. On the left are the approx-imate coefficients calculated from the table. On the right are the correct values(to five decimals), obtained with the aid of a table of the sine integral (see Ex-ercise 2). Figure 12 shows the difference between f (x) and F(x) (the sum of theFourier series using the approximate coefficients through a6). �

For hand calculation, choosing s to be a multiple of 4 makes many of thecosines “easy” numbers such as 1 and 0.5. When the calculation is done bydigital computer, this is not a consideration.

E X E R C I S E S

1. Since Table 3 gives sin(x)/x for seven points, seven cosine coefficients canbe calculated. Find a6.


Figure 12 Graph of the difference between f (x) = sin(x)/x and F(x), the sum ofthe Fourier series using the approximate coefficients a0 through a6.

2. Express the Fourier cosine coefficients of the example in terms of integralsof the form

Si((n + 1)π

) =∫ (n+1)π

0

sin(t)

tdt.

This is the sine integral function and is tabulated in many books, especiallyHandbook of Mathematical Functions, Abramowitz and Stegun, 1972.

3. Each entry in the list that follows represents the depth of the water in LakeOntario (minus the low-water datum of 242.8 feet) on the first of the cor-responding month. Assuming that the water level is a periodic functionof period one year, and that the observations are taken at equal intervals,compute the Fourier coefficients a0, a1, b1, a2, b2, thus identifying themean level, and fluctuations of period 12 months, 6 months, 4 months,and so forth. Take x0 as January, . . . , x11 as December, and x12 as Januaryagain.

Jan. 0.75 July 2.35Feb. 0.60 Aug. 2.15Mar. 0.65 Sept. 1.75Apr. 1.15 Oct. 1.05May 1.80 Nov. 1.00June 2.25 Dec. 0.90

4. The numbers in the table that follows represent the monthly precipi-tation (in inches of water) in Lake Placid, NY, averaged over the pe-riod 1950–1959. Find the approximate Fourier coefficients a0, . . . , a6 and

b1, . . . , b5.


Jan. 2.751 July 3.861Feb. 2.004 Aug. 4.088Mar. 3.166 Sept. 4.093Apr. 2.909 Oct. 3.434May 3.215 Nov. 2.902June 3.767 Dec. 3.011

1.9 Fourier IntegralIn Sections 1 and 2 of this chapter we developed the representation of a pe-riodic function in terms of sines and cosines with the same period. Then, bymeans of periodic extension, we obtained series representations for functionsdefined only on a finite interval. Now we must deal with nonperiodic functionsdefined for x between −∞ and ∞. Can such functions also be represented interms of sines and cosines? We make some transformations that suggest ananswer.

Suppose f (x) is defined for −∞ < x < ∞ and is sectionally smooth in everyfinite interval. Then for any positive a, f (x) can be represented in the interval−a < x < a by its Fourier series:

f (x) = a0 +∞∑

n=1

an cos

(nπx

a

)+ bn sin

(nπx

a

), −a < x < a,

a0 = 1

2a

∫ a

−af (x)dx, an = 1

a

∫ a

−af (x) cos

(nπx

a

)dx,

bn = 1

a

∫ a

−af (x) sin

(nπx

a

)dx. (1)

Example 1.Let

f (x) ={

e−x, 0 < x,0, x < 0.

For any a > 0, we have the Fourier series for f (x) on the interval −a < x < a:

f (x) = a0 +∞∑1

an cos

(nπx

a

)+ bn sin

(nπx

a

), −a < x < a,

a0 = 1 − e−a

2a, an = 1 − e−a cos(nπ)

a(1 + (nπ/a)2),

bn = (1 − e−a cos(nπ))nπ

a2(1 + (nπ/a)2). (2)

(The series converges to 1/2 at x = 0 and to e−a/2 at x = a.) �

1.9 Fourier Integral 107

Now we modify Eq. (1). Let λn = nπ/a and define two functions

Aa(λ) = 1

π

∫ a

−af (x) cos(λx)dx, Ba(λ) = 1

π

∫ a

−af (x) sin(λx)dx. (3)

Notice that

an = π

aAa(λn), bn = π

aBa(λn).

Because of this, the Fourier series Eq. (1) becomes

f (x) = a0 +∞∑

n=1

[Aa(λn) cos(λnx) + Ba(λn) sin(λnx)

] · λ, −a < x < a, (4)

where λ = π/a = λn+1 − λn.The form in which Eq. (4) is written is chosen to suggest an integral with

respect to λ over the interval 0 < λ < ∞. We may imagine a increasing toinfinity, so λ → 0 and

Aa(λ) → A(λ) = 1

π

∫ ∞

−∞f (x) cos(λx)dx, (5)

Ba(λ) → B(λ) = 1

π

∫ ∞

−∞f (x) sin(λx)dx, (6)

and a0 → 0. Then Eq. (4) suggests

f (x) =∫ ∞

0

[A(λ) cos(λx) + B(λ) sin(λx)

]dλ, −∞ < x < ∞. (7)

Example 1 (continued).For f (x) as in Example 1, we find

A(λ) = 1

π

∫ ∞

0e−x cos(λx)dx = 1

π(1 + λ2),

B(λ) = 1

π

∫ ∞

0e−x sin(λx)dx = λ

π(1 + λ2),

and therefore we expect that∫ ∞

0

[1

π(1 + λ2)cos(λx) + λ

π(1 + λ2)sin(λx)

]dx =

{e−x, 0 < x,0, x < 0.

�

The foregoing derivation is not a proof, but it does suggest the followingtheorem.


Fourier Integral Representation Theorem. Let f (x) be sectionally smooth onevery finite interval, and let

∫ ∞−∞ | f (x)|dx be finite. Then at every point x,

∫ ∞

0

(A(λ) cos(λx) + B(λ) sin(λx)

)dλ = 1

2

(f (x+) + f (x−)

),

−∞ < x < ∞, (8)

where

A(λ) = 1

π

∫ ∞

−∞f (x) cos(λx)dx, B(λ) = 1

π

∫ ∞

−∞f (x) sin(λx)dx. (9)

�

Equation (8) is called the Fourier integral representation of f (x); A(λ) and B(λ)

in Eq. (9) are the Fourier integral coefficient functions of f (x). The right-handside of Eq. (8) was also seen in the Fourier series convergence theorem. Sincef (x) is sectionally smooth, the expression in Eq. (8) is the same as f (x) almosteverywhere, so we often write Eq. (7) instead of Eq. (8).

Example 2.The function

f (x) ={

1, |x| < 1,0, |x| > 1

has the Fourier integral coefficient functions

A(λ) = 1

π

∫ ∞

−∞f (x) cos(λx)dx = 1

π

∫ 1

−1cos(λx)dx = 2 sin(λ)

πλ,

B(λ) = 0.

Since f (x) is sectionally smooth, the Fourier integral representation is legit-imate, and we write

f (x) =∫ ∞

0

2 sin(λ)

πλcos(λx)dλ.

(Actually the integral equals 12 at x = ±1, so equality is not strictly correct at

these two points.) �

Example 3.Find the Fourier integral representation of f (x) = exp(−|x|).Solution: Direct integration gives

A(λ) = 1

π

∫ ∞

−∞exp

(−|x|) cos(λx)dx, (10)


A(λ) = 2

π

∫ ∞

0e−x cos(λx)dx, (11)

A(λ) = 2

π

e−x(− cos(λx) + λ sin(λx))

1 + λ2

∣∣∣∣∞

0

= 2

π

1

1 + λ2. (12)

B(λ) = 0, because exp(−|x|) is even. Since exp(−|x|) is continuous and sec-tionally smooth, we may write

exp(−|x|) = 2

π

∫ ∞

0

cos(λx)

1 + λ2dλ, −∞ < x < ∞. �

These two examples illustrate the fact that, in general, one cannot evaluatethe integral in the Fourier integral representation. It is the theorem stated inthe preceding that allows us to write the equality between a suitable functionand its Fourier integral.

If f (x) is defined only in the interval 0 < x < ∞, one can construct an evenor odd extension whose Fourier integral contains only cos(λx) or sin(λx).These are called the Fourier cosine and sine integral representations of f , re-spectively.

Let f (x) be defined and sectionally smooth for 0 < x < ∞, and let∫ ∞0 | f (x)|dx < ∞. Then we write:

Fourier cosine integral representation

f (x) =∫ ∞

0A(λ) cos(λx)dλ, 0 < x < ∞

with A(λ) = 2

π

∫ ∞

0f (x) cos(λx)dx,

Fourier sine integral representation

f (x) =∫ ∞

0B(λ) sin(λx)dλ, 0 < x < ∞

with B(λ) = 2

π

∫ ∞

0f (x) sin(λx)dx.

Example 4.Find the Fourier sine and cosine integral representations of f (x) given for 0 < xby

f (x) ={

sin(x), 0 < x < π ,0, π < x.


Since f (x) = 0 for x > π , the integral for B(λ) reduces to one over the interval0 < x < π :

B(λ) = 2

π

∫ ∞

0f (x) sin(λx)dx = 2

π

∫ ∞

0sin(x) sin(λx)dx

= 2

π

[sin((λ − 1)x)

2(λ − 1)− sin((λ + 1)x)

2(λ + 1)

]π

0

= 2

π

[sin((λ − 1)π)

2(λ − 1)− sin((λ + 1)π)

2(λ + 1)

].

This expression can be simplified by using the fact that

sin((λ ± 1)π

) = sin(λπ ± π) = − sin(λπ).

Then, creating a common denominator, we obtain

B(λ) = −2 sin(λπ)

π(λ2 − 1).

Hence the Fourier sine integral representation of f (x) is

f (x) =∫ ∞

0

−2 sin(λπ)

π(λ2 − 1)sin(λx)dλ, 0 < x.

Since f (x) is continuous for 0 < x, the equality holds at every point.Similarly, we can compute the cosine coefficient function

A(λ) = −2(1 + cos(λπ))

π(λ2 − 1),

and the cosine integral representation of f (x) is

f (x) =∫ ∞

0

−2(1 + cos(λπ))

π(λ2 − 1)cos(λx)dλ, 0 < x.

Note that both A(λ) and B(λ) have removable discontinuities at λ = 1. �

It seems to be a rule of thumb that if the Fourier coefficient functions A(λ)

and B(λ) can be found in closed form for some function f (x), then the inte-gral in the Fourier integral representation cannot be carried out by elementarymeans, and vice versa. (See Exercise 3.)

Rules for operations on Fourier integrals generally follow the lines men-tioned in Section 1.5 for Fourier series. In particular: If f (x) is continuous andif both f (x) and f ′(x) have Fourier integral representations, then

f (x) =∫ ∞

0

[A(λ) cos(λx) + B(λ) sin(λx)

]dλ,


f ′(x) =∫ ∞

0

[−λA(λ) sin(λx) + λB(λ) cos(λx)]

dλ.

Example 5.Let f (x) = exp(−|x|) as in Example 3. Then its derivative is

f ′(x) ={−e−x, 0 < x,

ex, x < 0.

Clearly, f (x) is continuous, and both f (x) and f ′(x) have Fourier integral rep-resentations. The one for f (x) is in Example 3. Thus we have

f ′(x) =∫ ∞

0

2

π

−λ

1 + λ2sin(λx)dλ. �

E X E R C I S E S

1. Sketch the even and odd extensions of each of the following functions, andfind the Fourier cosine and sine integrals for f . Each function is given inthe interval 0 < x < ∞.

a. f (x) = e−x;

b. f (x) ={1, 0 < x < 1,

0, 1 < x;

c. f (x) ={

π − x, 0 < x < π ,0, π < x.

2. Find the Fourier integral representation of the following function ft(x).This is sometimes called a “window” because it is “open” for t − h < x <

t + h.

ft(x) ={

1, |x − t| < h,0, |x − t| > h.

3. Find the Fourier integral representation of each of the following functions.

a. f (x) = 1

1 + x2; b. f (x) = sin(x)

x.

(Hint: To evaluate the Fourier integral coefficient functions, consult theFourier integral representations found in the examples.)

4. In Exercise 3b, the integral∫ ∞−∞ | f (x)|dx is not finite. Nevertheless, A(λ)

and B(λ) do exist (B(λ) = 0). Find a rationale in the convergence theo-rem for saying that this function can be represented by its Fourier integral.(Hint: See Example 1.)

5. Find the Fourier integral representation of each of these functions:


a. f (x) ={

sin(x), −π < x < π ,0, |x| > π ;

b. f (x) ={

sin(x), 0 < x < π ,0, otherwise;

c. f (x) ={ |sin(x)|, −π < x < π ,

0, otherwise.

6. Show that if k and K are positive, then the following are true:

a.∫ ∞

0e−kx sin(x)dx = 1

1 + k2;

b.∫ ∞

0

1 − e−Kx

xsin(x)dx = tan−1(K);

c.∫ ∞

0

sin(x)

xdx = π

2.

(Part (a) by direct integration, (b) by integration of (a) with respect to kover the interval 0 to K , (c) by limit of (b) as K → ∞.)

7. Starting from Exercise 6c, show that

∫ ∞

0

sin(λz)

λdλ =

{π/2, 0 < z,0, z = 0,−π/2, z < 0.

Is this the Fourier integral of some function?

8. Change the variable of integration in the formulas for A and B, and justifyeach step of the following string of equalities. (Do not worry about chang-ing order of integration.)

f (x) = 1

π

∫ ∞

0

∫ ∞

−∞f (t)

(cos(λt) cos(λx) + sin(λt) sin(λx)

)dt dλ

= 1

π

∫ ∞

−∞f (t)

∫ ∞

0cos

(λ(t − x)

)dλdt

= 1

π

∫ ∞

−∞f (t)

[lim

ω→∞sin(ω(t − x))

t − x

]dt

= limω→∞

1

π

∫ ∞

−∞f (t)

sin(ω(t − x))

t − xdt.

The last integral is called Fourier’s single integral. Sketch the functionsin(ωv)/v as a function of v for several values of ω. What happens near

1.10 Complex Methods 113

v = 0? Sometimes notation is compressed and, instead of the last line, wewrite

f (x) =∫ ∞

−∞f (t)δ(t − x)dt.

Although δ is not, strictly speaking, a function, it is called Dirac’s delta func-tion.

1.10 Complex Methods

Fourier seriesSuppose that a function f (x) equals its Fourier series

f (x) = a0 +∞∑

n=1

an cos(nx) + bn sin(nx).

(We use period 2π for simplicity only.) A famous formula of Euler states that

eiθ = cos(θ) + i sin(θ), where i2 = −1.

Some simple algebra then gives the exponential definitions of the sine andcosine:

cos(θ) = 1

2

(eiθ + e−iθ

), sin(θ) = 1

2i

(eiθ − e−iθ

).

By substituting the exponential forms into the Fourier series of f we arrive atthe alternate form

f (x) = a0 + 1

2

∞∑n=1

an

(einx + e−inx

) − ibn

(einx − e−inx

)

= a0 + 1

2

∞∑n=1

(an − ibn)einx + (an + ibn)e−inx.

We are now led to define complex Fourier coefficients for f :

c0 = a0, cn = 1

2(an − ibn), c−n = 1

2(an + ibn), n = 1,2,3, . . . .

In terms of these two coefficients, we have

f (x) = c0 +∞∑

n=1

(cneinx + c−ne−inx

) =∞∑

−∞cneinx. (1)


This is the complex form of the Fourier series for f . It is easy to derive theuniversal formula

cn = 1

2π

∫ π

−π

f (x)e−inx dx, (2)

which is valid for all integers n, positive, negative, or zero. The complex form isused especially in physics and electrical engineering. Sometimes the functioncorresponding to a Fourier series can be recognized by use of the complexform.

Example.The series

∞∑n=1

(−1)n+1

ncos(nx)

may be considered the real part of

∞∑n=1

(−1)n+1

neinx =

∞∑n=1

(−1)n+1

n

(eix

)n(3)

because the real part of eiθ is cos(θ). The series on the right in Eq. (3) is recog-nized as a Taylor series,

∞∑n=1

(−1)n+1

n

(eix

)n = ln(1 + eix

).

Some manipulations yield

1 + eix = eix/2(eix/2 + e−ix/2

) = 2eix/2 cos

(x

2

),

ln(1 + eix

) = ix

2+ ln

(2 cos

(x

2

)).

The real part of ln(1+eix) is ln(2 cos(x/2)) when −π < x < π . Thus, we derivethe relation

ln

(2 cos

(x

2

))∼

∞∑n=1

(−1)n+1

ncos(nx), −π < x < π. (4)

(The series actually converges except at x = ±π,±3π, . . . .) �

1.10 Complex Methods 115

Fourier integralThe Fourier integral of a function f (x) defined in the entire interval −∞ <

x < ∞ can also be cast in complex form:

f (x) =∫ ∞

−∞C(λ)eiλx dλ. (5)

The complex Fourier integral coefficient function is given by

C(λ) = 1

2π

∫ ∞

−∞f (x)e−iλx dx. (6)

It is simple to show that

C(λ) = 1

2

(A(λ) − iB(λ)

), (7)

where A and B are the usual Fourier integral coefficients. The complex Fourierintegral coefficient is often called the Fourier transform of the function f (x).

Example.Find the complex Fourier integral representation of

f (x) ={

1, −a < x < a,0, x < |a|.

The coefficient function (or transform) of f is

C(λ) = 1

2π

∫ a

−ae−iλx dx = 1

2π

e−iλx

−iλ

∣∣∣∣a

−a

= 1

2π

eiλa − e−iλa

iλ= sin(λa)

πλ.

The representation of f is

f (x) =∫ ∞

−∞sin(λa)

πλeiλx dλ, −∞ < x < ∞.

Of course, at x = ±a, the integral converges to 1/2. �

This example brings out a fact about symmetry: If f (x) is even, C(λ) is real;if f (x) is odd, C(λ) is imaginary.

The Fourier integral or transform may be used to solve differential equationson the interval −∞ < x < ∞, in much the same way that Laplace transformis used.


E X E R C I S E S

1. Use the complex form

an − ibn = 1

π

∫ π

−π

f (x)e−inx dx, n �= 0,

to find the Fourier series of the function

f (x) = eαx, −π < x < π.

2. Find the complex Fourier series for the “square wave” with period 2π :

f (x) ={1, 0 < x < π ,

−1, −π < x < 0.

3. Find the complex Fourier integral representation of the following func-tions:

a. f (x) ={

e−x, x > 0,0, x < 0;

b. f (x) ={

sin(x), 0 < x < π ,0, elsewhere.

4. Find the complex Fourier integral for

a. f (x) ={

xe−x, 0 < x,0, x < 0;

b. f (x) = e−α|x| sin(x).

5. Relate the functions and series that follow by using complex form and Tay-lor series.

a. 1 +∞∑

n=1

rn cos(nx) = 1 − r cos(x)

1 − 2r cos(x) + r2, 0 ≤ r < 1;

b.∞∑

n=1

sin(nx)

n!= ecos(x) sin

(sin(x)

).

6. Show by integrating that∫ π

−π

einxe−imx dx ={

0, n �= m,2π, n = m,

and develop the formula for the complex Fourier coefficients using this ideaof orthogonality.

1.11 Applications of Fourier Series and Integrals 117

7. Find the function f (x) whose complex Fourier coefficient function isgiven.

a. C(λ) ={1, −1 < λ < 1,

0, otherwise;

b. C(λ) = e−|λ|.

8. Show that the complex Fourier coefficient of f (x) = e−x2is

C(λ) = e−λ2/4

2√

π.

Use a change of variable in the exponent. You need to know that∫ ∞

−∞e−z2

dz = √π.

1.11 Applications of Fourier Series and IntegralsFourier series and integrals are among the most basic tools of applied mathe-matics. In what follows, we give just a few applications that do not fall withinthe scope of the rest of this book.

A. Nonhomogeneous Differential EquationMany mechanical and electrical systems may be described by the differentialequation

y + αy + βy = f (t).

The function f (t) is called the “forcing function,” βy the “restoring term,” andαy the “damping term.” It is known (see Section 0.2) that: (1) a sine or cosinein f (t) will cause functions of the same period in y(t); (2) if f (t) is brokendown as a sum of simpler functions, y(t) can be broken down in the same way.

Suppose that f (t) is periodic with period 2π , and let its Fourier series be

f (t) = a0 +∞∑

n=1

an cos(nt) + bn sin(nt).

Then a particular solution y(t) will be periodic with period 2π ; it and its deriv-atives have Fourier series

y(t) = A0 +∞∑

n=1

An cos(nt) + Bn sin(nt),


y(t) =∞∑

n=1

−nAn sin(nt) + nBn cos(nt),

y(t) =∞∑

n=1

−n2An cos(nt) − n2Bn sin(nt).

Then the differential equation can be written in the form

βA0 +∞∑

n=1

(−n2An + αnBn + βAn

)cos(nt)

+∞∑

n=1

(−n2Bn − αnAn + βBn

)sin(nt) = a0 +

∞∑n=1

an cos(nt) + bn sin(nt).

The A’s and B’s are now determined by matching coefficients

βA0 = a0,(β − n2

)An + αnBn = an,

−αnAn + (β − n2

)Bn = bn.

When these equations are solved for the A’s and B’s, we find

An = (β − n2)an − αnbn

, Bn = (β − n2)bn + αnan

,

where

= (β − n2

)2 + α2n2.

Now, given the function f , the a’s and b’s can be determined, thus giving theA’s and B’s. The function y(t) represented by the series found is the periodicpart of the response. Depending on the initial conditions, there may also be atransient response, which dies out as t increases.

Example.Consider the differential equation

y + 0.4y + 1.04y = r(t).

If r(t) = sin(nt), the corresponding particular solution is

y(t) = −0.4n cos(nt) + (1.04 − n2) sin(nt)

(1.04 − n2)2 + (0.4n)2.


Next, suppose that r(t) is a square-wave function with Fourier series

r(t) =∞∑

n=1

2(1 − cos(nπ))

nπsin(nt).

The corresponding response is

y(t) =∞∑

n=1

2(1 − cos(nπ))

nπ· −0.4n cos(nt) + (1.04 − n2) sin(nt)

(1.04 − n2)2 + (0.4n)2.

Note that the term for n = 1 has a small denominator, causing a largeresponse. �

B. Boundary Value ProblemsBy way of introduction to the next chapter, we apply the idea of Fourier seriesto the solution of the boundary value problem

d2u

dx2+ pu = f (x), 0 < x < a,

u(0) = 0, u(a) = 0.

First, we will assume that f (x) is equal to its Fourier sine series,

f (x) =∞∑

n=1

bn sin

(nπx

a

), 0 < x < a.

And second, we will assume that the solution u(x), which we are seeking,equals its Fourier sine series,

u(x) =∞∑

n=1

Bn sin

(nπx

a

), 0 < x < a,

and that this series may be differentiated twice to give

d2u

dx2=

∞∑n=1

−(

n2π2

a2Bn

)sin

(nπx

a

), 0 < x < a.

When we insert the series forms for u, u′′, and f (x) into the differentialequation, we find that

∞∑n=1

(−n2π2

a2Bn + pBn

)sin

(nπx

a

)=

∞∑n=1

bn sin

(nπx

a

), 0 < x < a.


Since the coefficients of like terms in the two series must match, we may con-clude that (

p − n2π2

a2

)Bn = bn, n = 1,2,3, . . . .

If it should happen that p = m2π2/a2 for some positive integer m, there isno value of Bm that satisfies

(p − m2π2

a2

)Bm = bm

unless bm = 0 also, in which case any value of Bm is satisfactory. In summary,we may say that

Bn = bn

p − n2π2/a2

and

u(x) =∞∑

n=1

a2bn

a2p − n2π2sin

(nπx

a

),

with the agreement that a zero denominator must be handled separately.

Example.Consider the boundary value problem

d2u

dx2− u = −x, 0 < x < 1,

u(0) = 0, u(1) = 0.

We have found previously that

−x =∞∑

n=1

2(−1)n

πnsin(nπx), 0 < x < 1.

Thus, by the preceding development, the solution must be

u(x) =∞∑

n=1

2

π

(−1)n+1

n(n2π2 + 1)sin(nπx), 0 < x < 1.

Although this particular series belongs to a known function, one would not,in general, know any formula for the solution u(x) other than its Fourier sineseries. �


C. The Sampling TheoremOne of the most important results of information theory is the sampling the-orem, which is based on a combination of the Fourier series and the Fourierintegral in their complex forms. What the electrical engineer calls a signal isjust a function f (t) defined for all t. If the function is integrable, there is aFourier integral representation for it:

f (t) =∫ ∞

−∞C(ω) exp(iωt)dω,

C(ω) = 1

2π

∫ ∞

−∞f (t) exp(−iωt)dt.

A signal is called band limited if its Fourier transform is zero except in afinite interval, that is, if

C(ω) = 0, for |ω| > �.

Then � is called the cutoff frequency. If f is band limited, we can write it inthe form

f (t) =∫ �

−�

C(ω) exp(iωt)dω (1)

because C(ω) is zero outside the interval −� < ω < �. We focus our attentionon this interval by writing C(ω) as a Fourier series:

C(ω) =∞∑

−∞cn exp

(inπω

�

), −� < ω < �. (2)

The (complex) coefficients are

cn = 1

2�

∫ �

−�

C(ω) exp

(−inπω

�

)dω.

The point of the sampling theorem is to observe that the integral for cn ac-tually is a value of f (t) at a particular time. In fact, from the integral Eq. (1),we see that

cn = 1

2�f

(−nπ

�

).

Thus there is an easy way of finding the Fourier transform of a band-limitedfunction. We have

C(ω) = 1

2�

∞∑−∞

f

(−nπ

�

)exp

(inπω

�

)


= 1

2�

∞∑−∞

f

(nπ

�

)exp

(−inπω

�

), −� < ω < �.

By utilizing Eq. (1) again, we can reconstruct f (t):

f (t) =∫ �

−�

C(ω) exp(iωt)dω

= 1

2�

∞∑−∞

f

(nπ

�

)∫ �

−�

exp

(−inπω

�

)exp(iωt)dω.

Carrying out the integration and using the identity

sin(θ) = (eiθ − e−iθ )

2i,

we find

f (t) =∞∑

−∞f

(nπ

�

)sin(�t − nπ)

�t − nπ. (3)

This is the main result of the sampling theorem. It says that the band-limitedfunction f (t) may be reconstructed from the samples of f at t = 0, ±π/�, . . . .It is difficult to determine what functions are actually band limited. However,the process usually works quite well.

In practice, we must use a finite series to approximate the function

f (t) ∼=N∑

−N

f

(nπ

�

)sin(�t − nπ)

�t − nπ. (4)

Since the sampled values all come from the interval −Nπ/� to Nπ/�, theseries cannot attempt to approximate the function outside that interval. Ananimation on the CD shows the effects of choosing N and �.

Example.The function

f (t) = t2 + 2t

(1 + t2)2

is not band limited but can be approximated satisfactorily from a finite por-tion of the sum as in Eq. (4). Figure 13 shows results for N = 100 (thatis, 201 terms) and � = 4 and 10. The target function is dashed. Notice theimprovement. �


Figure 13 Graphs of approximation using sampling: Eq. (4) with N = 100 and� = 4 and 10.

E X E R C I S E S

1. Use the method of Part A to find a particular solution of

d2u

dt2+ 0.4

du

dt+ 1.04u = r(t),

where r(t) is periodic with period 4π and

r(t) = t

4π, 0 < t < 4π.

2. In the solution of Exercise 1, calculate the magnitude of the coefficients ofthe Fourier series of u(t) (periodic part).

3. A simply supported beam of length L has a point load w in the middle andaxial tension T. (See Exercises in Section 0.3.) Its displacement u(x) satisfiesthe boundary value problem

d2u

dx2− T

EIu = w

EIh(x), 0 < x < L,

u(0) = 0, u(L) = 0,

where h(x) is the “triangle function”

h(x) ={

2x/L, 0 < x < L/2,2(L − x)/L, L/2 < x < L.

Use the method of Part B to find u(x) as a sine series.

4. The inhomogeneity in the differential equation in Exercise 3 has a dis-continuous derivative. Find another way to solve the differential equation.Hint: Both u(x) and u′(x) must be continuous for 0 < x < L.


5. Use the software to approximate the function f (t) = e−t2by the Sampling

Theorem. Try � = 4, N = 2.

6. Simplify the final formula for sampling to

f (t) = sin(�t)∞∑

−∞f

(nπ

�

)(−1)n

�t − nπ.

1.12 Comments and References

The first use of trigonometric series occurred in the middle of the eighteenthcentury. Euler seems to have originated the use of orthogonality for the de-termination of coefficients. In the early nineteenth century Fourier made ex-tensive use of trigonometric series in studying problems of heat conduction(see Chapter 2). His claim, that an arbitrary function could be represented asa trigonometric series, led to an extensive reexamination of the foundationsof calculus. Fourier seems to have been among the first to recognize that afunction might have different analytical expressions in different places.

Dirichlet established sufficient conditions (similar to those of our conver-gence theorem) for the convergence of Fourier series around 1830. Later, Rie-mann was led to redefine the integral as part of his attempt to discover condi-tions on a function necessary and sufficient for the convergence of its Fourierseries. This problem has never been solved. Many other great mathematicianshave founded important theories (the theory of sets, for one) in the course ofstudying Fourier series, and they continue to be a subject of active research. Anentertaining and readable account of the history and uses of Fourier series isin The Mathematical Experience, by Davis and Hersh. (See the Bibliography.)

Historical interest aside, Fourier series and integrals are extremely impor-tant in applied mathematics, physics, and engineering, and they merit furtherstudy. A superbly written and organized book is Tolstov’s Fourier Series. Itsmathematical prerequisites are not too high. Fourier Series and Boundary ValueProblems by Churchill and Brown is a standard text for some engineering ap-plications.

About 1960 it became clear that the numerical computation of Fourier co-efficients could be rearranged to achieve dramatic reductions in the amountof arithmetic required. The result, called the fast Fourier transform, or FFT, hasrevolutionized the use of Fourier series in applications. See The Fast FourierTransform by James S. Walker.

The sampling theorem mentioned in the last section has become bread andbutter in communications engineering. For extensive information on this aswell as the FFT, see Integral and Discrete Transforms with Applications and ErrorAnalysis, by A.J. Jerri.


Chapter ReviewSee the CD for review questions.

Miscellaneous Exercises1. Find the Fourier sine series of the trapezoidal function given for

0 < x < π by

f (x) ={x/α, 0 < x < α,

1, α < x < π − α,(π − x)/α, π − α < x < π .

2. Show that the series found in Exercise 1 converges uniformly.

3. When α approaches 0, the function of Exercise 1 approaches a squarewave. Do the sine coefficients found in Exercise 1 approach those of asquare wave?

4. Find the Fourier cosine series of the function

F(x) =∫ x

0f (t)dt,

where f denotes the function in Exercise 1. Sketch.

5. Find the Fourier sine series of the function given in the interval 0 < x < aby the formula (α is a parameter between 0 and 1)

f (x) =

hx

αa, 0 < x < αa,

h(a − x)

(1 − α)a, αa < x < a.

6. Sketch the function of Exercise 5. To what does its Fourier sine seriesconverge at x = 0? at x = αa? at x = a?

7. Suppose that f (x) = 1, 0 < x < a. Sketch and find the Fourier series ofthe following extensions of f (x):

a. even extension;

b. odd extension;

c. periodic extension (period a);

d. even periodic extension;


e. odd periodic extension;

f. the one corresponding to f (x) = x, −a < x < 0.

8. Perform the same task as in Exercise 7, but f (x) = 0, 0 < x < a.

9. Find the Fourier series of the function given by

f (x) ={0, −a < x < 0,

2x, 0 < x < a.

Sketch the graph of f (x) and its periodic extension. To what values doesthe series converge at x = −a, x = −a/2, x = 0, x = a, and x = 2a?

10. Sketch the odd periodic extension and find the Fourier sine series of thefunction given by

f (x) ={

1, 0 < x < π2 ,

12 ,

π2 < x < π .

To what values does the series converge at x = 0, x = π/2, x = π , x =3π/2, and x = 2π ?

11. Sketch the even periodic extension of the function given in Exercise 10.Find its Fourier cosine series. To what values does the series converge atx = 0, x = π/2, x = π , x = 3π/2, and x = 2π ?

12. Find the Fourier cosine series of the function

g(x) ={1 − x, 0 < x < 1,

0, 1 < x < 2.

Sketch the graph of the sum of the cosine series.

13. Find the Fourier sine series of the function defined by f (x) = 1 − 2x,0 < x < 1. Sketch the graph of the odd periodic extension of f (x), anddetermine the sum of the sine series at points where the graph has ajump.

14. Following the same requirements as in Exercise 13, use the cosine seriesand the even periodic extension.

15. Find the Fourier series of the function given by

f (x) =

0, −π < x < −π2 ,

sin(2x), −π2 < x < π

2 ,0, π

2 < x < π .

Sketch the graph of the function.


16. Show that the function given by the formula f (x) = (π − x)/2, 0 < x <

2π , has the Fourier series

f (x) =∞∑1

sin(nx)

n, 0 < x < 2π.

Sketch f (x) and its periodic extension.

17. Use complex methods and a finite geometric series to show that

N∑n=1

cos(nx) = sin((N + 1

2 )x) − sin( 1

2 x)

2 sin( 12 x)

.

Then use trigonometric identities to identify

N∑n=1

cos(nx) = sin( 12 Nx) cos

(12 (N + 1)x

)sin( 1

2 x).

18. Identify the partial sums of the Fourier series in Exercise 16 as

SN(x) =N∑

n=1

sin(nx)

n.

The series of Exercise 17 is S′N(x). Use this information to locate the max-

ima and minima of SN(x) in the interval 0 ≤ x ≤ π . Find the value ofSN(x) at the first point in the interval 0 < x < π where S′

N(x) = 0 forN = 5. Compare to (π − x)/2 at that point.

19. Find the Fourier sine series of the function given by

f (x) ={

sin

(πx

a

), 0 < x < a,

0, a < x < π ,

assuming that 0 < a < π .

20. Find the Fourier cosine series of the function given in Exercise 19.

21. Find the Fourier integral representation of the function given by

f (x) ={1, 0 < x < a,

0, x < 0 or x > a.


22. Find the Fourier sine and cosine integral representations of the functiongiven by

f (x) ={ a − x

a, 0 < x < a,

0, a < x.

23. Find the Fourier sine integral representation of the function

f (x) ={

sin(x), 0 < x < π ,0, π < x.

24. Find the Fourier integral representation of the function

f (x) ={1/ε, α < x < α + ε,

0, elsewhere.

25. Use integration by parts to establish the equality∫ ∞

0e−λ cos(λx)dλ = 1

1 + x2.

26. The equation in Exercise 25 is valid for all x. Explain why its validityimplies that

2

π

∫ ∞

0

cos(λx)

1 + x2dx = e−λ, λ > 0.

27. Integrate both sides of the equality in Exercise 25 from 0 to t to derivethe equality ∫ ∞

0

e−λ sin(λt)

λdλ = tan−1(t).

28. Does the equality in Exercise 27 imply that

2

π

∫ ∞

0tan−1(t) sin(λt)dt = e−λ

λ?

29. From Exercise 27 derive the equality∫ ∞

0

1 − e−λ

λsin(λx)dλ = π

2− tan−1(x), x > 0.

30. Without using integration, obtain the Fourier series (period 2π) of eachof the following functions:


a. 2 + 4 sin(50x) − 12 cos(41x);

c. sin(4x + 2);

e. cos3(x);

b. sin2(5x);

d. sin(3x) cos(5x);

f. cos(2x + 13π).

31. Let the function f (x) be given in the interval 0 < x < 1 by the formula

f (x) = 1 − x.

Find (a) a sine series, (b) a cosine series, (c) a sine integral, and (d) acosine integral that equals the given function for 0 < x < 1. In each case,sketch the function to which the series or integral converges in the inter-val −2 < x < 2.

32. Verify the Fourier integral

∫ ∞

0cos(λq) exp

(−λ2t)dλ =

√π

4texp

(−q2

4t

), t > 0,

by transforming the left-hand side according to these steps: (a) Con-vert to an integral from −∞ to ∞ by using the evenness of the inte-grand; (b) replace cos(λq) by exp(iλq) (justify this step); (c) completethe square in the exponent; (d) change the variable of integration; (e) usethe equality ∫ ∞

−∞exp

(−u2)du = √

π.

33. Approximate the first seven cosine coefficients (a0, a1, . . . , a6) of thefunction

f (x) = 1

1 + x2, 0 < x < 1.

34. Use Fourier sine series representations of u(x) and of the function f (x) =x, 0 < x < a, to solve the boundary value problem

d2u

dx2− γ 2u = −x, 0 < x < a,

u(0) = 0, u(a) = 0.

35–43. For each of these exercises,

a. find the Fourier cosine series of the function;

b. determine the value to which the series converges at the given valuesof x;


c. sketch the even periodic extension of the given function for at leasttwo periods.


a. find the Fourier sine series of the function;

b. determine the value to which the series converges at the given valuesof x;

c. sketch the odd periodic extension of the given function for at leasttwo periods.

35. & 44. f (x) =

0, 0 < x <a

3,

x − a

3,

a

3< x <

2a

3, x = 0,

a

3,a,−a

2,

a

3,

2a

3< x < a.

36. & 45. f (x) =

1

2, 0 < x <

a

2, x = a

2,2a,0,−a,

1,a

2< x < a.

37. & 46. f (x) =

2x

a, 0 < x <

a

2, x = 0,

a

2,a,

3a

2,

(3a − 2x)

2a,

a

2< x < a.

38. & 47. f (x) =

x, 0 < x <a

2, x = 0,a,−a

2,

a

2,

a

2< x < a.

39. & 48. f (x) = (a − x)

a, 0 < x < a, x = 0,a,−a

2.

40. & 49. f (x) =

0, 0 < x <a

4,

1,a

4< x <

3a

4, x = 0,

a

4,

a

2,a,−3a

4,

0,3a

4< x < a.

41. & 50. f (x) = x(a − x), 0 < x < a, x = 0,−a,−a

2.

42. & 51. f (x) = ekx, 0 < x < a, x = 0,a

2,a,−a.


43. & 52. f (x) =

0, 0 < x <a

2, x = −a,

a

2,a,

1,a

2< x < a.


a. find the Fourier cosine integral representation of the function;

b. sketch the even extension of the function.


a. find the Fourier sine integral representation of the function;

b. sketch the odd extension of the function.

53. & 59. f (x) = e−x, 0 < x.

54. & 60. f (x) ={

e−x, 0 < x < a,0, a < x.

55. & 61. f (x) ={

1, 0 < x < b,0, b < x.

56. & 62. f (x) ={

cos(x), 0 < x < π ,0, π < x.

57. & 63. f (x) ={1 − x, 0 < x < 1,

0, 1 < x.

58. & 64. f (x) ={1, 0 < x < 1,

2 − x, 1 < x < 2,0, 2 < x.

65. (Cesaro summability.) Let f (x) be a periodic function with period 2π

whose Fourier coefficients are a0,a1,b1, . . . . Then, the partial sum

SN(x) = a0 +N∑

n=1

an cos(nx) + bn sin(nx)

is an approximation to f (x) if f is sectionally smooth and N is largeenough. The average of these approximations is

σN(x) = 1

N

(S1(x) + · · · + SN(x)

).

It is known that σN(x) converges uniformly to f (x) if f is continuous.Show that

σN(x) = a0 +N∑

n=1

N + 1 − n

N


).


66. In analogy to Lemma 2 of Section 7, prove that

N−1∑n=0

sin

((n + 1

2

)y

)= sin2( 1

2 Ny)

sin( 12 y)

.

67. Following the lines of Section 7, show that

σN(x) − f (x) = 1

2Nπ

∫ π

−π

[f (x + y) − f (x)

]( sin( 12 Ny)

sin( 12 y)

)2

dy.

This equality is the key to the proof of uniform convergence mentionedin Exercise 65.

68. In a study of river freezing, E.P. Foltyn and H.T. Shen [St. Lawrence Riverfreeze-up forecast, Journal of Waterway, Port, Coastal and Ocean Engi-neering, 112 (1986): 467–481] use data spanning 33 years to find thisFourier series representation of the air temperature in Massena, NY:

T(t) = a0 +∞∑

n=1

an cos(2nπ t) + bn sin(2nπ t).

Here T is temperature in ◦C, t is time in years, and the origin is Oct. 1.The first coefficients were found to be

a0 = 6.638, a1 = 5.870, b1 = −13.094, a2 = 0.166, b2 = 0.583,

and the remaining coefficients were all less than 0.3 in absolute value.The authors decided to exclude all the terms from a2 and b2 up, so theirapproximation could be written

T(t) ∼= a0 + A sin(2π t + θ).

a. Find the average temperature in Massena.

b. Find A, the amplitude of the annual variation, and the phase angle θ .

c. Find the approximate date when the minimum temperature occurs.

d. Find the dates when the approximate temperature passes through 0.

e. Discuss the effect on the answer to part d if the next two terms of theseries were included.

69. In each part that follows, a function is equated to its Fourier series asjustified by the Theorem of Section 3. By evaluating both sides of theequality at an appropriate value of x, derive the second equality.


a. |x| = 1

2− 4

π2

∞∑k=0

1

(2k + 1)2cos

((2k + 1)πx

), −1 < x < 1,

π2

8= 1 + 1

9+ 1

25+ · · · ;

b.4

π

∞∑k=0

1

2k + 1sin

((2k + 1)πx

) ={1, 0 < x < 1,

−1, −1 < x < 0,

π

4= 1 − 1

3+ 1

5− 1

7+ · · · ;

c. |sin(x)| = 2

π− 4

π

∞∑n=1

1

4n2 − 1cos(2nx),

1

2= 1

3+ 1

15+ 1

35+ · · · .

The Heat EquationC H A P T E R

2

2.1 Derivation and Boundary ConditionsAs the first example of the derivation of a partial differential equation, weconsider the problem of describing the temperature in a rod or bar of heat-conducting material. In order to simplify the problem as much as possible, weshall assume that the rod has a uniform cross section (like an extrusion) andthat the temperature does not vary from point to point on a section. Thus, ifwe use a coordinate system as suggested in Fig. 1, we may say that the temper-ature depends only on position x and time t.

The basic idea in developing the partial differential equation is to apply thelaws of physics to a small piece of the rod. Specifically, we apply the law ofconservation of energy to a slice of the rod that lies between x and x + x(Fig. 2).

The law of conservation of energy states that the amount of heat that entersa region plus what is generated inside is equal to the amount of heat that leavesplus the amount stored. The law is equally valid in terms of rates per unit timeinstead of amounts.

Now let q(x, t) be the heat flux at point x and time t. The dimensions of qare1 [q] = H/tL2, and q is taken to be positive when heat flows to the right.The rate at which heat is entering the slice through the surface at x is Aq(x, t),where A is the area of a cross section. The rate at which heat is leaving the slicethrough the surface at x + x is Aq(x + x, t).

1Square brackets are used to symbolize “dimension of.” H = heat energy, t = time, T =temperature, L = length, m = mass, and so forth.

135

136 Chapter 2 The Heat Equation

Figure 1 Rod of heat-conducting material.

Figure 2 Slice cut from rod.

The rate of heat storage in the slice of material is proportional to the rate ofchange of temperature. Thus, if ρ is the density and c is the heat capacity perunit mass ([c] = H/mT), we may approximate the rate of heat storage in theslice by

ρcAx∂u

∂t(x, t),

where u(x, t) is the temperature.There are other ways in which heat may enter (or leave) the section of rod

we are looking at. One possibility is that heat is transferred by radiation orconvection from (or to) a surrounding medium. Another is that heat is con-verted from another form of energy — for instance, by resistance to an elec-trical current or by chemical or nuclear reaction. All of these possibilities welump together in a “generation rate.” If the rate of generation per unit volumeis g, [g] = H/tL3, then the rate at which heat is generated in the slice is Ax g.(Note that g may depend on x, t, and even u.)

We have now quantified the law of conservation of energy for the slice ofrod in the form

Aq(x, t) + Ax g = Aq(x + x, t) + Ax ρc∂u

∂t. (1)

After some algebraic manipulation, we have

q(x, t) − q(x + x, t)

x+ g = ρc

∂u

∂t.

The ratio

q(x + x, t) − q(x, t)

x

Chapter 2 The Heat Equation 137

should be recognized as a difference quotient. If we allow x to decrease, thisquotient becomes, in the limit,

limx→0

q(x + x, t) − q(x, t)

x= ∂q

∂x.

The limit process thus leaves the law of conservation of energy in the form

−∂q

∂x+ g = ρc

∂u

∂t. (2)

We are not finished, since there are two dependent variables, q and u, in thisequation. We need another equation relating q and u. This relation is Fourier’slaw of heat conduction, which in one dimension may be written

q = −κ∂u

∂x.

In words, heat flows downhill (q is positive when ∂u/∂x is negative) at a rateproportional to the gradient of the temperature. The proportionality factor κ ,called the thermal conductivity, may depend on x if the rod is not uniform andalso may depend on temperature. However, we will usually assume it to be aconstant.

Substituting Fourier’s law in the heat balance equation yields

∂

∂x

(κ

∂u

∂x

)+ g = ρc

∂u

∂t. (3)

Note that κ , ρ, and c may all be functions. If, however, they are independentof x, t, and u, we may write

∂2u

∂x2+ g

κ= ρc

κ

∂u

∂t. (4)

The equation is applicable where the rod is located and after the experimentstarts: for 0 < x < a and for t > 0. The quantity κ/ρc is often written as kand is called the thermal diffusivity. Table 1 shows approximate values of theseconstants for several materials.

For some time we will be working with the heat equation without genera-tion,

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (5)

which, to review, is supposed to describe the temperature u in a rod of lengtha with uniform properties and cross section, in which no heat is generated andwhose cylindrical surface is insulated.

Some qualitative features can be obtained from the partial differential equa-tion itself. Suppose that u(x, t) satisfies the heat equation, and imagine a graph


c ρ κ k = κ

ρc

Material(

calg ◦C

) ( gcm3

) (cal

s cm ◦C

) (cm2

s

)Aluminum 0.21 2.7 0.48 0.83Copper 0.094 8.9 0.92 1.1Steel 0.11 7.8 0.11 0.13Glass 0.15 2.6 0.0014 0.0036Concrete 0.16 2.3 0.0041 0.011Ice 0.48 0.92 0.004 0.009

Table 1 Typical values of constants

of u(x, t∗), with t∗ a fixed time. If a portion of the graph is shaped like U, J orbackwards J, the graph is concave there — that is, ∂2u/∂x2 is positive. Then bythe heat equation, ∂u/∂t must be positive as well. Vice versa, when the graphis convex, ∂2u/∂x2 and hence ∂u/∂t must be negative. Thus, a solution of theheat equation tends to straighten out.

This equation alone is not enough information to completely specify thetemperature, however. Each of the functions

u(x, t) = x2 + 2kt,

u(x, t) = e−kt sin(x)

satisfies the partial differential equation, and so do their sum and difference.Clearly this is not a satisfactory situation either from the mathematical or

physical viewpoint; we would like the temperature to be uniquely determined.More conditions must be placed on the function u. The appropriate additionalconditions are those that describe the initial temperature distribution in therod and what is happening at the ends of the rod.

The initial condition is described mathematically as

u(x,0) = f (x), 0 < x < a,

where f (x) is a given function of x alone. In this way, we specify the initialtemperature at every point of the rod.

The boundary conditions may take a variety of forms. First, the temperatureat either end may be held constant, for instance, by exposing the end to an ice-water bath or to condensing steam. We can describe such conditions by theequations

u(0, t) = T0, u(a, t) = T1, t > 0,

where T0 and T1 may be the same or different. More generally, the temperatureat the boundary may be controlled in some way, without being held constant.If x0 symbolizes an endpoint, the condition is

u(x0, t) = α(t), (6)


where α is a function of time. Of course, the case of a constant function isincluded here. This type of boundary condition is called a Dirichlet conditionor condition of the first kind.

Another possibility is that the heat flow rate is controlled. Since Fourier’slaw associates the heat flow rate and the gradient of the temperature, we canwrite

∂u

∂x(x0, t) = β(t), (7)

where β is a function of time. This is called a Neumann condition or conditionof the second kind. We most frequently take β(t) to be identically zero. Thenthe condition

∂u

∂x(x0, t) = 0

corresponds to an insulated surface, for this equation says that the heat flow iszero.

Still another possible boundary condition is

c1u(x0, t) + c2∂u

∂x(x0, t) = γ (t), (8)

called third kind or a Robin condition. This kind of condition can also be real-ized physically. If the surface at x = a is exposed to air or other fluid, then theheat conducted up to that surface from inside the rod is carried away by con-vection. Newton’s law of cooling says that the rate at which heat is transferredfrom the body to the fluid is proportional to the difference in temperaturebetween the body and the fluid. In symbols, we have

q(a, t) = h(u(a, t) − T(t)

), (9)

where T(t) is the air temperature. After application of Fourier’s law, this be-comes

−κ∂u

∂x(a, t) = hu(a, t) − hT(t). (10)

This equation can be put into the form of Eq. (8). (Note: h is called the con-vection coefficient or heat transfer coefficient; [h] = H/L2tT.)

All of the boundary conditions given in Eqs. (6), (7), and (8) involve thefunction u and/or its derivative at one point. If more than one point is in-volved, the boundary condition is called mixed. For example, if a uniform rodis bent into a ring and the ends x = 0 and x = a are joined, appropriate bound-ary conditions would be


u(0, t) = u(a, t), t > 0, (11)

∂u

∂x(0, t) = ∂u

∂x(a, t), t > 0, (12)

both of mixed type.Many other kinds of boundary conditions exist and are even realizable, but

the four kinds already mentioned here are the most commonly encountered.An important feature common to all four types is that they involve a linearoperation on the function u.

The heat equation, an initial condition, and a boundary condition for eachend form what is called an initial value–boundary value problem. For instance,one possible problem would be

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (13)

u(0, t) = T0, 0 < t, (14)

−κ∂u

∂x(a, t) = h

(u(a, t) − T1

), 0 < t, (15)

u(x,0) = f (x), 0 < x < a. (16)

Notice that the boundary conditions may be of different kinds at differentends.

Although we shall not prove it, it is true that there is one, and only one,solution to a complete initial value–boundary value problem.

We have derived the heat equation (4) as a mathematical model for the tem-perature in a “rod,” suggesting an object that is much longer than it is wide.The equation applies equally well to a “slab,” an object that is much wider thanit is thick. The important feature is that we may assume in either case that thetemperature varies in only one space direction (along the length of the rodor the thickness of the slab). In Chapter 5, we derive a multidimensional heatequation.

It may come as a surprise that the partial differential equations of this sec-tion have another completely different but equally important physical inter-pretation. Suppose that a static medium occupies a region of space betweenx = 0 and x = a (a slab!) and that we wish to study the concentration u, mea-sured in units of mass per unit volume, of another substance, whose moleculesor atoms can move, or diffuse, through the medium. We assume that the con-centration is a function of x and t only and designate q(x, t) to be the massflux ([q] = m/tL2). Then the principle of conservation of mass may be appliedto a layer of the medium between x and x + x to obtain the equation

q(x, t) + x g = q(x + x, t) + x∂u

∂t(x, t). (17)


When we rearrange Eq. (17) and take the limit as x approaches 0, it becomes

−∂q

∂x+ g = ∂u

∂t. (18)

In these equations, g is a “generation rate” ([g] = m/tL3), a function that ac-counts for any gain or loss of the substance from the layer by means other thanmovement in the x-direction. For example, the substance may participate in achemical reaction with the medium at a rate proportional to its concentration(a first-order reaction) so that in this case the generation rate is

g = −ku(x, t). (19)

The concentration and the mass flux are linked by a phenomenological re-lation called Fick’s first law, written in one dimension as

q = −D∂u

∂x. (20)

In words, the diffusing substance moves toward regions of lower concentra-tion at a rate proportional to the gradient of the concentration. The coefficientof proportionality D, usually constant, is called the diffusivity. By combiningFick’s law with Eq. (18) arising from the conservation of mass, we obtain thediffusion equation

∂2u

∂x2+ g

D= 1

D

∂u

∂t. (21)

At a boundary of the medium, the concentration of the diffusing substancemay be controlled, leading to a condition like Eq. (6), or the flux of the sub-stance may be controlled, leading via Fick’s law to a condition like Eq. (7); animpermeable surface corresponds to zero flux. If a boundary is covered with apermeable film, then the flux through the film is usually taken to be propor-tional to the difference in concentrations on the two sides of the film. Supposethat the surface in question is at x = a. Then these statements may be expressedsymbolically as

q(a, t) = h(u(a, t) − C(t)

), (22)

where the proportionality constant h is called the film coefficient and C is theconcentration outside the medium. Using Fick’s law here leads to the equation

−D∂u

∂x(a, t) = hu(a, t) − hC(t). (23)

This equation is analogous to Eq. (10) and can be put into the form of Eq. (8).


E X E R C I S E S

1. Give a physical interpretation for the problem in Eqs. (13)–(16).

2. Verify that the following functions are solutions of the heat equation (5):

u(x, t) = exp(−λ2kt

)cos(λx),

u(x, t) = exp(−λ2kt

)sin(λx).

3. Suppose that the rod exchanges heat through the cylindrical surface by con-vection with a surrounding fluid at temperature U (constant). Newton’s lawof cooling says that the rate of heat transfer is proportional to exposed areaand temperature difference. What is g in Eq. (1)? What form does Eq. (4)take?

4. Suppose that the end of the rod at x = 0 is immersed in an insulated con-tainer of water or other fluid; that the temperature of the fluid is the sameas the temperature of the end of the rod; that the heat capacity of the fluidis C units of heat per degree. Show that this situation is represented math-ematically by the equation

C∂u

∂t(0, t) = κA

∂u

∂x(0, t),

where A is the cross-sectional area of the rod.

5. Put Eq. (10) into Eq. (8) form. Notice that the signs still indicate that heatflows in the direction of lower temperature. That is, if u(a, t) > T(t), thenq(a, t) is positive and the gradient of u is negative. Show that, if the surfaceat x = 0 (left end) is exposed to convection, the boundary condition wouldread

κ∂u

∂x(0, t) = hu(0, t) − hT(t).

Explain the signs.

6. Suppose the surface at x = a is exposed to radiation. The Stefan–Boltzmannlaw of radiation says that the rate of radiation heat transfer is proportionalto the difference of the fourth powers of the absolute temperatures of thebodies:

q(a, t) = σ(u4(a, t) − T4

).

Use this equation and Fourier’s law to obtain a boundary condition forradiation at x = a to a body at temperature T.

7. The difference cited in Exercise 6 may be written

u4 − T4 = (u − T)(u3 + u2T + uT2 + T3

).

2.2 Steady-State Temperatures 143

Under what conditions might the second factor on the right be taken ap-proximately constant? If the factor were constant, the boundary conditionwould be linear.

8. Interpret this problem in terms of diffusion. Be sure to explain how theboundary conditions could arise physically.

D∂2u

∂x2+ K = ∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = C,∂u

∂x(a, t) = 0, 0 < t,

u(x,0) = 0, 0 < x < a.

2.2 Steady-State TemperaturesBefore tackling a complete heat conduction problem, we shall solve a simpli-fied version called the steady-state or equilibrium problem. We begin with thisproblem:

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (1)

u(0, t) = T0, 0 < t, (2)

u(a, t) = T1, 0 < t, (3)

u(x,0) = f (x), 0 < x < a. (4)

We may think of u(x, t) as the temperature in a cylindrical rod, with insulatedlateral surface, whose ends are held at constant temperatures T0 and T1.

Experience indicates that after a long time under the same conditions, thevariation of temperature with time dies away. In terms of the function u(x, t)that represents temperature, we thus expect that the limit of u(x, t), as t tendsto infinity, exists and depends only on x,

limt→∞ u(x, t) = v(x),

and also that

limt→∞

∂u

∂t= 0.

The function v(x), called the steady-state temperature distribution, must stillsatisfy the boundary conditions and the heat equation, which are valid for allt > 0.


Example.For the preceding problem, v(x) should be the solution to the problem

d2v

dx2= 0, 0 < x < a, (5)

v(0) = T0, v(a) = T1. (6)

On integrating the differential equation twice, we find

dv

dx= A, v(x) = Ax + B.

The constants A and B are to be chosen so that v(x) satisfies the boundaryconditions:

v(0) = B = T0, v(a) = Aa + B = T1.

When the two equations are solved for A and B, the steady-state distributionbecomes

v(x) = T0 + (T1 − T0)x

a. (7)

Of course, Eqs. (5) and (6), which together form the steady-state problemcorresponding to Eqs. (1)–(4), could have been derived from scratch, as wasdone in Chapter 0, Section 3. Here, however, we see it as part of a more com-prehensive problem. �

We can establish this rule for setting up the steady-state problem corre-sponding to a given heat conduction problem: Take limits in all equations thatare valid for large t (the partial differential equation and the boundary condi-tions), replacing u and its derivatives with respect to x by v and its derivatives,and replacing ∂u/∂t by 0.

Example.Find the steady-state problem and solution of Eqs. (13)–(16) of Section 2.1,which were

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (8)

u(0, t) = T0, 0 < t, (9)

−κ∂u

∂x(a, t) = h

(u(a, t) − T1

), 0 < t, (10)

u(x,0) = f (x), 0 < x < a. (11)


When the rule given here is applied to this problem, we are led to the followingequations:

d2v

dx2= 0, 0 < x < a,

v(0) = T0, −κv′(a) = h(v(a) − T1

).

The solution of the differential equation is v(x) = A + Bx. The boundary con-ditions require that A and B satisfy

v(0) = T0: A = T0,

−κv′(a) = h(v(a) − T1

): −κB = h(A + Ba − T1).

Solving simultaneously, we find

A = T0, B = h(T1 − T0)

κ + ha.

Thus the steady-state solution of Eqs. (8)–(11) is

v(x) = T0 + xh(T1 − T0)

κ + ha. (12)

�

In both of these examples, the steady-state temperature distribution hasbeen uniquely determined by the differential equation and boundary condi-tions. This is usually the case, but not always.

Example.For the problem

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (13)

∂u

∂x(0, t) = 0, 0 < t, (14)

∂u

∂x(a, t) = 0, 0 < t, (15)

u(x,0) = f (x), 0 < x < a, (16)

which describes the temperature in an insulated rod that also has insulatedends, the corresponding steady-state problem for v(x) = limt→∞ u(x, t) is

d2v

dx2= 0, 0 < x < a,

dv

dx(0) = 0,

dv

dx(a) = 0.


It is easy to see that v(x) = T (any constant) is a solution to this problem.However, there is no information to tell what value T should take. Thus, thisboundary value problem has infinitely many solutions. �

It should not be supposed that every steady-state temperature distributionhas a straight-line graph. This is certainly not the case in the problem of Exer-cise 1.

While the steady-state solution gives us some valuable information aboutthe solution of an initial value–boundary value problem, it also is importantas the first step in finding the complete solution. We now isolate the “rest” ofthe unknown temperature u(x, t) by defining the transient temperature distri-bution,

w(x, t) = u(x, t) − v(x).

The name transient is appropriate because, according to our assumptionsabout the behavior of u for large values of t, we expect w(x, t) to tend to zeroas t tends to infinity.

In general, the transient also satisfies an initial value–boundary value prob-lem that is similar to the original one but is distinguished by having a homo-geneous partial differential equation and boundary conditions. To illustratethis point, we shall treat the problem stated in Eqs. (1)–(4) whose steady-statesolution is given by Eq. (7).

By using the equality u(x, t) = w(x, t)+ v(x) and what we know about v —that is, Eqs. (5) and (6) — we make the original problem for u(x, t) into a newproblem for w(x, t), as shown in what follows.

∂2u

∂x2= ∂2w

∂x2+ d2v

dx2by rules of calculus,

= ∂2w

∂x2because of Eq. (5),

∂u

∂t= ∂w

∂t+ dv

dtby rules of calculus,

= ∂w

∂tv(x) does not depend on t,

∂2w

∂x2= 1

k

∂w

∂tby substituting into Eq. (1),

u(0, t) = w(0, t) + v(0) by definition of the transient,

T0 = w(0, t) + T0 from Eqs. (2) and (6),

u(a, t) = w(a, t) + v(a) by definition of the transient,

T1 = w(a, t) + T1 from Eqs. (3) and (6),

u(x,0) = w(x,0) + v(x) by definition of the transient,

f (x) = w(x,0) + T0 + (T1 − T0)x/a from Eqs. (4) and (7).


Now we collect and simplify these transformations of Eqs. (1)–(4) to get aninitial value–boundary value problem for w:

∂2w

∂x2= 1

k

∂w

∂t, 0 < x < a, 0 < t, (17)

w(0, t) = 0, 0 < t, (18)

w(a, t) = 0, 0 < t, (19)

w(x,0) = f (x) −[

T0 + (T1 − T0)x

a

](20)

≡ g(x), 0 < x < a. (21)

In the last line, we have just renamed the combination of f (x) and v(x) inEq. (20).

In the next section, we shall see how the problem for the transient temper-ature can be solved. The mathematical purpose of setting up the steady-stateproblem and then the transient problem is that the transient problem is ho-mogeneous. You can test this by trying w(x, t) ≡ 0: This function satisfies thepartial differential equation (17) and the boundary conditions (18) and (19). Itis crucially important for the method we will develop to have a homogeneouspartial differential equation and boundary conditions.

E X E R C I S E S

See extra exercises on the CD.1. State and solve the steady-state problem corresponding to

∂2u

∂x2− γ 2(u − U) = 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0, u(a, t) = T1, 0 < t,

u(x,0) = 0, 0 < x < a.

Also find a physical interpretation of this problem. (See Exercise 3, Sec-tion 1.)

2. State the problem satisfied by the transient temperature distribution corre-sponding to the problem in Exercise 1.

3. Obtain the steady-state solution of the problem

∂2u

∂x2+ γ 2(u − T) = 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T, u(a, t) = T, 0 < t,

u(x,0) = T1x

a, 0 < x < a.


Can you think of a physical interpretation of this problem? Note the differ-ence between the partial differential equation in this exercise and in Exer-cise 1. What happens if γ = π/a?

4. State the initial value–boundary value problem satisfied by the transienttemperature distribution corresponding to Eqs. (8)–(11).

5. Find the steady-state solution of the problem

∂

∂x

(κ

∂u

∂x

)= cρ

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0, u(a, t) = T1, 0 < t

if the conductivity varies in a linear fashion with x: κ(x) = κ0 + βx, whereκ0 and β are constants.

6. Find and sketch the steady-state solution of

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t

together with boundary conditions

a.∂u

∂x(0, t) = 0, u(a, t) = T0;

b. u(0, t) − ∂u

∂x(0, t) = T0,

∂u

∂x(a, t) = 0;

c. u(0, t) − ∂u

∂x(0, t) = T0, u(a, t) + ∂u

∂x(a, t) = T1.

7. Find the steady-state solution of this problem, where r is a constant thatrepresents heat generation.

∂2u

∂x2+ r = 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0,∂u

∂x(a, t) = 0, 0 < t.

8. Find the steady-state solution of

∂2u

∂x2+ γ 2

(U(x) − u

) = 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = U0,∂u

∂x(a, t) = 0, 0 < t,

where U(x) = U0 + Sx (U0,S are constants).

2.3 Example: Fixed End Temperatures 149

9. This problem describes the diffusion of a substance in a medium that ismoving with speed S to the right. The unknown function u(x, t) is the con-centration of the diffusing substance. Write out the steady-state problemand solve it. (D, U , and S are constants.)

D∂2u

∂x2= ∂u

∂t+ S

∂u

∂x, 0 < x < a, 0 < t,

u(0, t) = U, u(a, t) = 0, 0 < t,

u(x,0) = 0, 0 < x < a.

2.3 Example: Fixed End Temperatures

In Section 1 we saw that the temperature u(x, t) in a uniform rod with insu-lated material surface would be determined by the problem

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (1)

u(0, t) = T0, 0 < t, (2)

u(a, t) = T1, 0 < t, (3)

u(x,0) = f (x), 0 < x < a (4)

if the ends of the rod are held at fixed temperatures and if the initial tempera-ture distribution is f (x). In Section 2 we found that the steady-state tempera-ture distribution,

v(x) = limt→∞ u(x, t),

satisfied the boundary value problem

d2v

dx2= 0, 0 < x < a, (5)

v(0) = T0, v(a) = T1. (6)

In fact, we were able to find v(x) explicitly:

v(x) = T0 + (T1 − T0)x

a. (7)

We also defined the transient temperature distribution as

w(x, t) = u(x, t) − v(x)


and determined that w satisfies the boundary value–initial value problem

∂2w

∂x2= 1

k

∂w

∂t, 0 < x < a, 0 < t, (8)

w(0, t) = 0, 0 < t, (9)

w(a, t) = 0, 0 < t, (10)

w(x,0) = f (x) − v(x) ≡ g(x), 0 < x < a. (11)

Our objective is to determine the transient temperature distribution, w(x, t),and — since v(x) is already known — the unknown temperature will be

u(x, t) = v(x) + w(x, t). (12)

The problem in w can be attacked by a method called product method, sepa-ration of variables, or Fourier’s method. For this method to work, it is essentialto have homogeneous partial differential equation and boundary conditions.Thus, the method may be applied to the transient distribution w but not tothe original function u. Of course, because both the partial differential equa-tion and the boundary conditions satisfied by w(x, t) are homogeneous, thefunction w ≡ 0 satisfies them. Because this solution itself is obvious and is ofno help in satisfying the initial condition, it is called the trivial solution. Weare seeking the unobvious, nontrivial solutions, so we shall avoid the trivialsolution at every turn.

The general idea of the method is to assume that the solution of the par-tial differential equation has the form of a product: w(x, t) = φ(x)T(t). Werequire that neither of the factors φ(x) and T(t) be identically 0, since thatwould lead back to the trivial solution. Now, each of the factors depends ononly one variable, so we have

∂2w

∂x2= φ′′(x)T(t),

∂w

∂t= φ(x)T ′(t).

The partial differential equation becomes

φ′′(x)T(t) = 1

kφ(x)T ′(t),

and on dividing through by φT we find

φ′′(x)

φ(x)= T ′(t)

kT(t), 0 < x < a, 0 < t.

Here is the key argument: The ratio on the left contains functions of x aloneand cannot vary with t. On the other hand, the ratio on the right containsfunctions of t alone and cannot vary with x. Since this equality must hold for


all x in the interval 0 < x < a and for all t > 0, the common value of the twosides must be a constant, varying neither with x nor t:

φ′′(x)

φ(x)= p,

T ′(t)

kT(t)= p.

Now we have two ordinary differential equations for the two factor functions:

φ′′ − pφ = 0, T ′ − pkT = 0. (13)

The two boundary conditions on w may also be stated in the product form:

w(0, t) = φ(0)T(t) = 0, w(a, t) = φ(a)T(t) = 0.

There are two ways these equations can be satisfied for all t > 0. Either thefunction T(t) ≡ 0 for all t, which is forbidden, or the other factors must bezero. Therefore, we have

φ(0) = 0, φ(a) = 0. (14)

Our job now is to solve Eqs. (13) and satisfy the boundary conditions (14)while avoiding the trivial solution.

Case 1: If p > 0, the solutions of Eqs. (13) are

φ(x) = c1 cosh(√

px) + c2 sinh

(√px

), T(t) = cepkt .

Now we apply the boundary conditions:

φ(0) = 0: c1 = 0,

φ(a) = 0: c2 sinh(√

pa) = 0.

Because the sinh function is 0 only when its argument is 0 — clearly not trueof

√pa — we have c1 = c2 = 0 and φ(x) ≡ 0, which is not acceptable.

Case 2: If we take p = 0, the solutions of the differential equations (13) areφ(x) = c1 + c2x, T(t) = c. The boundary conditions require

φ(0) = 0: c1 = 0,

φ(a) = 0: c2a = 0.

Again we have φ(x) ≡ 0.Case 3: We now try a negative constant. Replacing p by −λ2 in Eqs. (13)

gives us the two equations

φ′′ + λ2φ = 0, T ′ + λ2kT = 0,

whose solutions are

φ(x) = c1 cos(λx) + c2 sin(λx), T(t) = c exp(−λ2kt

).


If φ has the form given in the preceding, the boundary conditions require thatφ(0) = c1 = 0, leaving φ(x) = c2 sin(λx). Then φ(a) = c2 sin(λa) = 0.

We now have two choices: either c2 = 0, making φ(x) ≡ 0 for all values of x,or sin(λa) = 0. We reject the first possibility, for it leads to the trivial solutionw(x, t) ≡ 0. In order for the second possibility to hold, we must have λ =nπ/a, where n = ±1,±2,±3, . . . . The negative values of n do not give anynew functions, because sin(−θ) = − sin(θ). Hence we allow n = 1,2,3, . . .

only. We shall set λn = nπ/a.Incidentally, because the differential equations (13) and the boundary con-

ditions (14) for φ(x) are homogeneous, any constant multiple of a solution isstill a solution. We shall therefore remember this fact and drop the constant c2

in φ(x). Likewise, we delete the c in T(t).To review our position, we have, for each n = 1,2,3, . . . , a function φn(x) =

sin(λnx) and an associated function Tn(t) = exp(−λ2nkt). The product wn(x, t)

= sin(λnx) exp(−λ2nkt) has these properties:

1. ∂2wn

∂x2 = −λ2nwn; ∂wn

∂t = −λ2nkwn; and therefore wn satisfies the heat equa-

tion.

2. wn(0, t) = sin(0)e−λ2nkt = 0 for any n and t; and therefore wn satisfies the

boundary condition at x = 0.

3. wn(a, t) = sin(λna)e−λ2nkt = 0 for any n and t because λna = nπ and

sin(nπ) = 0. Therefore wn satisfies the boundary condition at x = a.

Now we call on the Principle of Superposition in order to continue.

Principle of Superposition.If u1,u2, . . . are solutions of the same linear, homogeneous equations, then sois

u = c1u1 + c2u2 + · · · . �

In fact, we have infinitely many solutions, so we need an infinite series tocombine them all:

w(x, t) =∞∑

n=1

bn sin(λnx) exp(−λ2

nkt). (15)

Using an infinite series brings up questions about convergence that we are go-ing to ignore. However, it is easy to verify that the function defined by theseries does satisfy the boundary conditions: At x = 0 and at x = a, each termis 0, so the sum is 0 as well. To check the partial differential equation, we haveto differentiate w(x, t) by differentiating each term of the series. This done, itis easy to see that terms match and the heat equation is satisfied.


Notice that the choice of the coefficients bn does not enter into the check-ing of the partial differential equation and the boundary conditions. Thus,Eq. (15) plays the role of a general solution of Eqs. (8)–(10).

Of the four parts of the original problem, only the initial condition has notyet been satisfied. At t = 0, the exponentials in Eq. (15) are all unity. Thus theinitial condition takes the form

w(x,0) =∞∑

n=1

bn sin

(nπx

a

)= g(x), 0 < x < a. (16)

We immediately recognize a problem in Fourier series, which is solved bychoosing the constants bn according to the formula

bn = 2

a

∫ a

0g(x) sin

(nπx

a

)dx. (17)

If the function g is continuous and sectionally smooth, we know that theFourier series actually converges to g(x) in the interval 0 < x < a, so the solu-tion that we have found for w(x, t) actually satisfies all requirements set on w.Even if g does not satisfy these conditions, it can be shown that the solutionwe have arrived at is the best we can do.

Once the transient temperature has been determined, we find the originalunknown u(x, t) as the sum of the transient and the steady-state solutions,

u(x, t) = v(x) + w(x, t).

Example.Suppose the original problem to be

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0, 0 < t,

u(a, t) = T1, 0 < t,

u(x,0) = 0, 0 < x < a.

The steady-state solution is

v(x) = T0 + (T1 − T0)x

a.

The transient temperature, w(x, t) = u(x, t) − v(x), satisfies

∂2w

∂x2= 1

k

∂w

∂t, 0 < x < a, 0 < t,

w(0, t) = 0, 0 < t,


w(a, t) = 0, 0 < t,

w(x,0) = −T0 − (T1 − T0)x

a≡ g(x), 0 < x < a.

According to the preceding calculations, w has the form

w(x, t) =∞∑

n=1


nkt)

(18)

and the initial condition is

w(x,0) =∞∑

n=1

bn sin

(nπx

a

)= g(x), 0 < x < a.

The coefficients bn are given by

bn = 2

a

∫ a

0

[−T0 − (T1 − T0)

x

a

]sin

(nπx

a

)dx

= 2T0

a

cos(nπx/a)

(nπ/a)

∣∣∣∣a

0

− 2

a2(T1 − T0)

sin(nπx/a) − (nπx/a) cos(nπx/a)

(nπ/a)2

∣∣∣∣a

0

= −2T0

nπ

(1 − (−1)n

) + 2(T1 − T0)

nπ(−1)n

bn = −2

nπ

(T0 − T1(−1)n

).

Now the complete solution (see Fig. 3) is

u(x, t) = w(x, t) + T0 + (T1 − T0)x

a,

where

w(x, t) = − 2

π

∞∑n=1

T0 − T1(−1)n

nsin(λnx) exp

(−λ2nkt

). (19)

The solution of this problem is shown as an animation on the CD. �

We can discover certain features of u(x, t) by examining the solution. First,u(x,0) really is zero (0 < x < a) because the Fourier series converges to −v(x)at t = 0. Second, when t is positive but very small, the series for w(x, t) willalmost equal −T0 − (T1 − T0)x/a. But at x = 0 and x = a, the series addsup to zero (and w(x, t) is a continuous function of x); thus u(x, t) satisfiesthe boundary conditions. Third, when t is large, exp(−λ2

1kt) is small, and the


Figure 3 The solution of the example with T1 = 100 and T0 = 20. The functionu(x, t) is graphed as a function of x for four values of t, chosen so that the dimen-sionless time kt/a2 has the values 0.001, 0.01, 0.1, and 1. For kt/a2 = 1, the steadystate is practically achieved. See the CD.

other exponentials are still smaller. Then w(x, t) may be well approximatedby the first term (or first few terms) of the series. Finally, as t → ∞, w(x, t)disappears completely.

E X E R C I S E S

Also see Separation of Variables Step by Step on the CD.

1. Write out the first few terms of the series for w(x, t) in Eq. (19).

2. If k = 1 cm2/s, a = 1 cm, show that after t = 0.5 s the other terms of theseries for w are negligible compared with the first term. Sketch u(x, t) fort = 0, t = 0.5, t = 1.0, and t = ∞. Take T0 = 100, T1 = 300.

3. We can see from Eq. (19) that the dimensionless combinations x/a andkt/a2 appear in the sine and exponential functions. Reformulate the partialdifferential equation (8) in terms of the dimensionless variables. ξ = x/a,τ = kt/a2. Set u(x, t) = U(ξ, τ ).

4. Sketch the functions φ1, φ2, and φ3, and verify that they satisfy the bound-ary conditions φ(0) = 0, φ(a) = 0.

In Exercises 5–8, solve the problem

∂2w

∂x2= 1

k

∂w

∂t, 0 < x < a, 0 < t,

w(0, t) = 0, w(a, t) = 0, 0 < t,

w(x,0) = g(x), 0 < x < a

for the given function g(x).


5. g(x) = T0 (constant).

6. g(x) = βx (β is constant).

7. g(x) = β(a − x) (β is constant).

8. g(x) =

2T0x

a, 0 < x <

a

2,

2T0(a − x)

a,

a

2< x < a.

9. A.N. Virkar, T.B. Jackson, and R.A. Cutler [Thermodynamic and kinetic ef-fects of oxygen removal on the thermal conductivity of aluminum nitride,Journal of the American Ceramic Society, 72 (1989): 2031–2042] use the fol-lowing boundary value problem to study the kinetics of oxygen removalfrom a grain of aluminum nitride by diffusion:

∂C

∂t= D

∂2C

∂x2, 0 < x < a, 0 < t,

C(0, t) = C1, C(a, t) = C1, 0 < t,

C(x,0) = C0, 0 < x < a.

In these equations, C is the oxygen concentration, D is the diffusion con-stant, a is the thickness of a grain, C0 and C1 are known concentrations.

a. Find the steady-state solution, v(x).

b. State the problem (partial differential equation, boundary conditionsand initial condition) for the transient, w(x, t) = C(x, t) − v(x).

c. Solve the problem for w(x, t), and write out the complete solutionC(x, t).

d. The concentration in the center of the grain, C(a/2, t), varies from C0

at time t = 0 toward C1 as t increases. Suppose we want to find out howlong it takes for this concentration to complete 90% of the change it willmake from C0 to C1; that is, we want to solve this equation for t:

C

(a

2, t

)− C0 = 0.9(C1 − C0).

Show that this equation is equivalent to the equation

w

(a

2, t

)= −0.1(C1 − C0).

Find an approximate formula for the solution by using just the first termof the series for w(x, t).

2.4 Example: Insulated Bar 157

e. Use the formula in d to find t explicitly for a = 5 × 10−6 m, D =10−11 cm2/s. Be careful to check dimensions.

2.4 Example: Insulated BarWe shall consider again the uniform bar that was discussed in Section 1. Letus suppose now that the ends of the bar at x = 0 and x = a are insulated in-stead of being held at constant temperatures. The boundary value–initial valueproblem that describes the temperature in this rod is:

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (1)

∂u

∂x(0, t) = 0,

∂u

∂x(a, t) = 0, 0 < t, (2)

u(x,0) = f (x), 0 < x < a, (3)

where f (x) is supposed to be a given function.We saw in Section 2 that the solution of the steady-state problem is not

unique. However, the mathematical purpose behind finding the steady-statesolution is to pave the way for a homogeneous problem (partial differentialequation and boundary conditions) for the transient. In this example the par-tial differential equation and boundary conditions are already homogeneous.Thus, we do not need the steady-state solution or the transient problem. Wemay look for u(x, t) directly.

Assume that u has the product form u(x, t) = φ(x)T(t), with neither factoridentically 0. The heat equation becomes

φ′′(x)T(t) = 1

kφ(x)T ′(t),

and the variables are separated by dividing through by φT, leaving

φ′′(x)

φ(x)= T ′(x)

kT(t), 0 < x < a, 0 < t.

In order that a function of x equal a function of t, their mutual value mustbe a constant. If that constant were positive, T would be an increasing expo-nential function of time, which would be unacceptable. It is also easy to showthat if the constant were positive, φ could not satisfy the boundary conditionswithout being identically zero.

Assuming then a negative constant, we can write

φ′′(x)

φ(x)= −λ2 = T ′(t)

kT(t)


and separate these equalities into two ordinary differential equations linked bythe common parameter λ2:

φ′′ + λ2φ = 0, 0 < x < a, (4)

T ′ + λ2kT = 0, 0 < t. (5)

The boundary conditions on u can be translated into conditions on φ, be-cause they are homogeneous conditions. The boundary conditions in productform are

∂u

∂x(0, t) = φ′(0)T(t) = 0, 0 < t,

∂u

∂x(a, t) = φ′(a)T(t) = 0, 0 < t.

To satisfy these equations, we must have the function T(t) always zero (whichwould make u(x, t) ≡ 0), or else

φ′(0) = 0, φ′(a) = 0.

The second alternative avoids the trivial solution.We now have a homogeneous differential equation for φ together with ho-

mogeneous boundary conditions:

φ′′ + λ2φ = 0, 0 < x < a, (6)

φ′(0) = 0, φ′(a) = 0. (7)

A problem of this kind is called an eigenvalue problem. We are looking for thosevalues of the parameter λ2 for which nonzero solutions of Eqs. (6) and (7)may exist. Those values are called eigenvalues, and the corresponding solutionsare called eigenfunctions. Note that the significant parameter is λ2, not λ. Thesquare is used only for convenience. It is worth mentioning that we alreadysaw an eigenvalue problem in Section 3 and in the Euler buckling problem ofChapter 0.

The general solution of the differential equation in Eq. (6) is

φ(x) = c1 cos(λx) + c2 sin(λx).

Applying the boundary condition at x = 0, we see that φ′(0) = c2λ = 0, giv-ing c2 = 0 or λ = 0. We put aside the case λ = 0 and assume c2 = 0, soφ(x) = c1 cos(λx). Then the second boundary condition requires that φ′(a) =−c1λ sin(λa) = 0. Once again, we may have c1 = 0 or sin(λa) = 0. But c1 = 0makes φ(x) ≡ 0. We choose therefore to make sin(λa) = 0 by restricting λ to


the values π/a,2π/a,3π/a, . . . . We label the eigenvalues with a subscript:

λ2n =

(nπ

a

)2

, φn(x) = cos(λnx), n = 1,2, . . . .

Notice that any constant multiple of an eigenfunction is still an eigenfunction;thus, we may take c1 = 1 for simplicity.

Returning to the case λ = 0, we see that Eqs. (6) and (7) become

φ′′ = 0, 0 < x < a,

φ′(0) = 0, φ′(a) = 0.

The solution of the differential equation is φ(x) = c1 + c2x. Both boundaryconditions say c2 = 0. Therefore φ(x) is a (any) constant. Thus 0 is an eigen-value of the problem Eqs. (6) and (7), and we designate

λ20 = 0, φ0(x) = 1.

Let us summarize our findings by saying that the eigenvalue problem,Eqs. (6) and (7), has the solution

λ2

0 = 0, φ0(x) = 1,

λ2n =

(nπ

a

)2

, φn(x) = cos(λnx), n = 1,2, . . . .

Now that the numbers λ2n are known, we can solve Eq. (5) for T(t), finding

T0(t) = 1, Tn(t) = exp(−λ2

nkt).

The products φn(x)Tn(t) give solutions of the partial differential equation (1)that satisfy the boundary conditions, Eq. (2):

u0(x, t) = 1, un(x, t) = cos(λnx) exp(−λ2

nkt). (8)

Because the partial differential equation and the boundary conditions are alllinear and homogeneous, the principle of superposition applies, and any linearcombination of solutions is also a solution. The solution u(x, t) of the wholesystem may therefore have the form

u(x, t) = a0 +∞∑1

an cos(λnx) exp(−λ2

nkt). (9)

There is only one condition of the original set remaining to be satisfied, theinitial condition Eq. (3). For u(x, t) in the form of Eq. (9), the initial condi-tion is

u(x,0) = a0 +∞∑1

an cos(λnx) = f (x), 0 < x < a.


Figure 4 The solution of the example, u(x, t), as a function of x for several times.The initial temperature distribution is f (x) = T0 + (T1 − T0)x/a. For this illustra-tion, T0 = 20,T1 = 100, and the times are chosen so that the dimensionless timekt/a2 takes the values 0.001, 0.01, 0.1, and 1. The last case is indistinguishable fromthe steady state. See the CD also.

Because λn = nπ/a, we recognize a problem in Fourier series and can imme-diately cite formulas for the coefficients:

a0 = 1

a

∫ a

0f (x)dx, an = 2

a

∫ a

0f (x) cos

(nπx

a

)dx. (10)

When these coefficients are computed and substituted in the formulas foru(x, t), that function becomes the solution to the initial value–boundary valueproblems, Eqs. (1)–(3). Notice that when t → ∞, all other terms in the sum-mation for u(x, t) disappear, leaving

limt→∞ u(x, t) = a0 = 1

a

∫ a

0f (x)dx.

Example.Find the complete solution of Eqs. (1)–(3) for the initial temperature dis-tribution f (x) = T0 + (T1 − T0)x/a. It requires no integration to find thata0 = (T1 + T0)/2. The remaining coefficients are

an = 2

a

∫ a

0

(T0 + (T1 − T0)x

a

)cos

(nπx

a

)dx

= 2(T1 − T0)cos(nπ) − 1

n2π2.

Thus the solution is given by Eq. (9) with these coefficients for a0 and an.A graph of u(x, t) as a function of x is shown in Fig. 4 and as an animation onthe CD. �


E X E R C I S E S

1. Using the initial condition

u(x,0) = T1x

a, 0 < x < a,

find the solution u(x, t) of Eqs. (1)–(3). Sketch u(x,0), u(x, t) for somet > 0 (using the first three terms of the series), and the steady-state solu-tion.

2. Repeat Exercise 1 using the initial condition

u(x,0) = T0 + T1

(x

a

)2

, 0 < x < a.

3. Same as Exercise 1, but with initial condition

u(x,0) =

2T0x

a, 0 < x <

a

2,

2T0(a − x)

a,

a

2< x < a.

4. Solve Eqs. (1)–(3) using the initial condition u(x,0) = f (x), where

f (x) =

T1, 0 < x <a

2,

T2,a

2< x < a.

5. Consider this heat problem, which is related to Eqs. (1)–(3):

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

∂u

∂x(0, t) = S0,

∂u

∂x(a, t) = S1, 0 < t,

u(x,0) = f (x), 0 < x < a.

a. Show that the steady-state problem has a solution if and only if S0 = S1,and give a physical reason why this should be true. (Recall that the heatflux q is proportional to the derivative of u with respect to x.) Find thesteady-state solution if this condition is met.

b. If the steady-state solution v(x) exists, show that the “transient,”w(x, t) = u(x, t) − v(x), has the boundary conditions

∂w

∂x(0, t) = 0,

∂w

∂x(a, t) = 0, 0 < t.


c. Show that the function u(x, t) = A(kt + x2/2) + Bx satisfies the heatequation for arbitrary A and B and that A and B can be chosen to satisfythe boundary conditions

∂u

∂x(0, t) = S0,

∂u

∂x(a, t) = S1, 0 < t.

What happens to u(x, t) as t increases if S0 �= S1?

6. Verify that un(x, t) in Eq. (8) satisfies the partial differential equation (1)and the boundary conditions, Eq. (2).

7. State the eigenvalue problem associated with the solution of the heat prob-lem in Section 3. Also state its solution.

8. Suppose that the function φ(x) satisfies the relation

φ′′(x)

φ(x)= p2 > 0.

Show that the boundary conditions φ′(0) = 0, φ′(a) = 0, then force φ(x)to be identically 0. Thus, a positive “separation constant” can only lead tothe trivial solution.

9. Refer to Eqs. (9) and (10), which give the solution of the problem statedin Eqs. (1)–(3). If f is sectionally continuous, the coefficients an → 0 asn → ∞. For t = t1 > 0, fixed, the solution is

u(x, t1) = a0 +∞∑1

an exp(−λ2

nkt1

)cos(λnx)

and the coefficients of this cosine series are

An(t1) = an exp(−λ2

nkt1

).

Show that An(t1) → 0 so rapidly as n → ∞ that the series given in thepreceding converges uniformly 0 ≤ x ≤ a. (See Chapter 1, Section 4, The-orem 1.) Show the same for the series that represents

∂2u

∂x2(x, t1).

10. Sketch the functions φ1, φ2, and φ3 and verify graphically that they satisfythe boundary conditions of Eq. (7).

11. The boundary conditions Eq. (2) require that

∂u

∂x(0, t) = 0, 0 < t,

and similarly at x = a. Does this mean that u is constant at x = 0?

2.5 Example: Different Boundary Conditions 163

12. This table gives values of u(0, t) for the function u found in the exampleand shown in Fig. 4. Make a graph of u(0, t) and describe the graph inwords.

kt/a2: 0.001 0.003 0.01 0.03 0.1 0.3 1u(0, t): 22.9 24.9 29.0 35.6 47.9 58.3 60.0

13. Check that the partial differential equation and boundary conditions aresatisfied by the series in Eq. (9).

2.5 Example: Different Boundary ConditionsIn many important cases, boundary conditions at the two endpoints will bedifferent kinds. In this section we shall solve the problem of finding the tem-perature in a rod having one end insulated and the other held at a constanttemperature. The boundary value–initial value problem satisfied by the tem-perature in the rod is

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (1)

u(0, t) = T0, 0 < t, (2)

∂u

∂x(a, t) = 0, 0 < t, (3)

u(x,0) = f (x), 0 < x < a. (4)

It is easy to verify that the steady-state solution of this problem is v(x) = T0.Using this information, we can find the boundary value–initial value problemsatisfied by the transient temperature w(x, t) = u(x, t) − T0:

∂2w

∂x2= 1

k

∂w

∂t, 0 < x < a, 0 < t, (5)

w(0, t) = 0,∂w

∂x(a, t) = 0, 0 < t, (6)

w(x,0) = f (x) − T0 = g(x), 0 < x < a. (7)

Since this problem is homogeneous, we can attack it by the method of sep-aration of variables. The assumption that w(x, t) has the form of a product,w(x, t) = φ(x)T(t), and insertion of w in that form into the partial differentialequation (5) lead, as before, to

φ′′(x)

φ(x)= T ′(t)

kT(t)= constant. (8)


The boundary conditions take the form

φ(0)T(t) = 0, 0 < t, (9)

φ′(a)T(t) = 0, 0 < t. (10)

As before, we conclude that φ(0) and φ′(a) should both be zero:

φ(0) = 0, φ′(a) = 0. (11)

By trial and error we find that a positive or zero separation constant in Eq. (8)forces φ(x) ≡ 0. Thus we take the constant to be −λ2. The separated equationsare

φ′′ + λ2φ = 0, 0 < x < a, (12)

T ′ + λ2kT = 0, 0 < t. (13)

Now, the general solution of the differential equation (12) is


The boundary condition, φ(0) = 0, requires that c1 = 0, leaving

φ(x) = c2 sin(λx).

The boundary condition at x = a now takes the form

φ′(a) = c2λ cos(λa) = 0.

The three choices are: c2 = 0, which gives the trivial solution; λ = 0, whichshould be investigated separately (Exercise 2), and cos(λa) = 0. The third al-ternative — the only acceptable one — requires that λa be an odd multiple ofπ/2, which we may express as

λn = (2n − 1)π

2a, n = 1,2, . . . . (14)

Thus, we have found that the eigenvalue problem consisting of Eqs. (11)and (12) has the solution

λn = (2n − 1)π

2a, φn(x) = sin(λnx), n = 1,2,3, . . . . (15)

With the eigenfunctions and eigenvalues now in hand, we return to the dif-ferential equation (13), whose solution is

Tn(t) = exp(−λ2

nkt).


As in previous cases, we assemble the general solution of the homogeneousproblem expressed in Eqs. (5)–(7) by forming a general linear combination ofour product solutions,

w(x, t) =∞∑

n=1


nkt). (16)

The choice of the coefficients, bn, must be made so as to satisfy the initial con-dition, Eq. (8). Using the form of w given by Eq. (16), we find that the initialcondition is

w(x,0) =∞∑

n=1

bn sin

((2n − 1)πx

2a

)= g(x), 0 < x < a. (17)

A routine Fourier sine series for the interval 0 < x < a would involve the func-tions sin(nπx/a), rather than the functions we have. By one of several means(Exercises 10–12), it may be shown that the series in Eq. (17) represents thefunction g(x), provided that g is sectionally smooth and that we choose thecoefficients by the formula

bn = 2

a

∫ a

0g(x) sin

((2n − 1)πx

2a

)dx. (18)

Now the original problem is completely solved. The solution is

u(x, t) = T0 +∞∑

n=1


nkt). (19)

It should be noted carefully that the T0 term in Eq. (19) is the steady-statesolution in this case; it is not part of the separation-of-variables solution.

Example.Find the solution of Eqs. (1)–(4) with the initial condition

u(x,0) = T1, 0 < x < a.

Then g(x) = T1 −T0, 0 < x < a, and the coefficients as determined by Eq. (18)are

bn = (T1 − T0)4

π(2n − 1).

Therefore, the complete solution of the boundary value–initial value problemwith initial condition u(x,0) = T1 would be

u(x, t) = T0 + (T1 − T0)4

π

∞∑n=1

1

2n − 1sin(λnx) exp

(−λ2nkt

). (20)

See Fig. 5 for graphs and an animation on the CD. �


Figure 5 Solution of the example, Eq. (20): u(x, t) is shown as a function of xfor various times, which are chosen so that the dimensionless time kt/a2 takes thevalues 0.001, 0.01, 0.1, 1.0. For the illustration, T0 has been chosen equal to 0 andT1 = 100.

Now that we have been through three major examples, we can outline themethod we have been using to solve linear boundary value–initial value prob-lems. Up to this moment we have seen only homogeneous partial differentialequations, but a nonhomogeneity that is independent of t can be treated bythe same technique.

Summary of Separation of VariablesPrepareIf the partial differential equation or a boundary condition or both are nothomogeneous, first find a function v(x), independent of t, that satisfies thepartial differential equation and the boundary conditions. Since v(x) does notdepend on t, the partial differential equation applied to v(x) becomes an ordi-nary differential equation. Finding v(x) is just a matter of solving a two-pointboundary value problem.

Determine the initial value–boundary value problem satisfied by the “tran-sient solution” w(x, t) = u(x, t) − v(x). This must be a homogeneous problem.That is, the partial differential equation and the boundary conditions (but notusually the initial condition) are satisfied by the constant function 0.

SeparateAssuming that w(x, t) = φ(x)T(t), with neither factor 0, separate the partialdifferential equation into two ordinary differential equations, one for φ(x) andone for T(t), linked by the separation constant, −λ2. Reduce the boundaryconditions to conditions on φ alone.


SolveSolve the eigenvalue problem for φ. That is, find the values of λ2 for whichthe eigenvalue problem has nonzero solutions. Label the eigenfunctions andeigenvalues φn(x) and λ2

n.Solve the ordinary differential equation for the time factors, Tn(t).

Combine and Satisfy Remaining ConditionForm the general solution of the homogeneous problem as a sum of constantmultiples of the product solutions:

w(x, t) =∑

cnφn(x)Tn(t).

Choose the cn so that the initial condition is satisfied. This may or may notbe a routine Fourier series problem. If not, an orthogonality principle mustbe used to determine the coefficients. (We shall see the theory in Sections 7and 8.)

CheckForm the solution of the original problem

u(x, t) = v(x) + w(x, t)

and check that all conditions are satisfied.

E X E R C I S E S

See Common Eigenvalue Problems on the CD.

1. Find the steady-state solution of the problem stated in Eqs. (1)–(4).

2. Determine whether 0 is an eigenvalue of the eigenvalue problem statedin Eqs. (11) and (12). That is, take λ = 0 and see whether the solution isnonzero.

3. Solve the problem stated in Eqs. (1)–(4), taking f (x) = Tx/a.

4. Solve the problem stated in Eqs. (1)–(4) if

f (x) ={

T0, 0 < x < a/2,T1, a/2 < x < a.

5. Solve the nonhomogeneous problem

∂2u

∂x2= 1

k

∂u

∂t− T

a2, 0 < x < a, 0 < t,


u(0, t) = T0,∂u

∂x(a, t) = 0, 0 < t,

u(x,0) = T0, 0 < x < a.

6. Solve this problem for the temperature in a rod in contact along the lateralsurface with a medium at temperature 0.

∂2u

∂x2= 1

k

∂u

∂t+ γ 2u, 0 < x < a, 0 < t,

u(0, t) = 0,∂u

∂x(a, t) = 0, 0 < t,

u(x,0) = T0, 0 < x < a.

7. Solve the problem

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

∂u

∂x(0, t) = 0, u(a, t) = T0, 0 < t,

u(x,0) = T1, 0 < x < a.

8. Compare the solution of Exercise 7 with Eq. (20). Can one be turned intothe other?

9. Solve the problem in Exercise 7 taking T0 = 0 and

u(x,0) = T1 cos

(πx

2a

), 0 < x < a.

10. a. Show that the eigenfunctions found in this section are orthogonal. Thatis, prove that ∫ a

0sin(λnx) sin(λmx)dx =

{0 (m �= n),a

2(m = n)

when λn = (2n−1)π

2a .

b. Use the orthogonality relation in part a to justify the formula inEq. (18).

11. To justify the expansion of Eq. (17), for an arbitrary sectionally smoothg(x),

∞∑n=1

bn sin

((2n − 1)πx

2a

)= g(x), 0 < x < a,


construct the function G(x) with these properties:

G(x) = g(x), 0 < x < a,

G(x) = g(2a − x), a < x < 2a.

Show that G(x) corresponds to the series

G(x) ∼∞∑

N=1

BN sin

(Nπx

2a

), 0 < x < 2a.

12. Show that the BN of the series in the preceding equation satisfy

BN = 0 (N even), BN = 2

a

∫ a

0g(x) sin

(Nπx

2a

)dx (N odd).

13. a. Solve this problem over the interval 0 < x < 2a.

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < 2a, 0 < t,

u(0, t) = T0, u(2a, t) = T0, 0 < t,

u(x,0) = g(x), 0 < x < 2a.

A function f is given over the interval 0 < x < a, and g is an extensionof f defined by

g(x) ={

f (x), 0 < x < a,f (2a − x), a < x < 2a.

b. Explain why the solution of the problem comprising Eqs. (1)–(4) isexactly the same as the solution of the problem in part a.

14. In the ceramics industry, the following problem has to be analyzed forparameter measurement. A cylindrical rod of uniform porous material issuspended vertically so that its lower end is immersed in water. The cylin-drical surface and the upper end are sealed — with wax, for example. Theconcentration of water in the rod (weight per unit volume) is a functionC(x, t) that satisfies the boundary value problem

D∂2C

∂x2= ∂C

∂t, 0 < x < L, 0 < t,

C(0, t) = C0,∂C

∂x(L, t) = 0, 0 < t,

C(x,0) = 0, 0 < x < L.


In these equations, D is the diffusion constant, L is the length of the cylin-der, and C0 is the saturation concentration, which depends on the porosityof the material. Find C(x, t).

15. Use the solution of Exercise 14 to find an expression for the total weightof water absorbed by the rod,

W(t) = A

∫ L

0C(x, t)dx.

16. A plot of W(t)/C0 as a function of s = √Dt/L2 over the range 0 to 2

resembles a slanted line segment joined by a curve to a horizontal linesegment. The slope of the slanted line segment in this graph is approx-imately 1. Experimenters plot measured values of W(t)/C0 vs

√t to get

a similar graph. Then they use the slope of the slanted line segment tofind D. Explain how.

2.6 Example: ConvectionWe have seen three examples in which boundary conditions specified eitheru or ∂u/∂x. Now we shall study a case where a condition of the third kindis involved. The physical model is conduction of heat in a rod with insulatedlateral surface whose left end is held at constant temperature and whose rightend is exposed to convective heat transfer. The boundary value–initial valueproblem satisfied by the temperature in the rod is

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t, (1)

u(0, t) = T0, 0 < t, (2)

−κ∂u

∂x(a, t) = h

(u(a, t) − T1

), 0 < t, (3)

u(x,0) = f (x), 0 < x < a. (4)

We found in Section 2 that the steady-state solution of this problem is

v(x) = T0 + xh(T1 − T0)

κ + ha. (5)

Now, since the original boundary conditions were nonhomogeneous, weform the problem for the transient solution w(x, t) = u(x, t) − v(x). By directsubstitution it is found that

∂2w

∂x2= 1

k

∂w

∂t, 0 < x < a, 0 < t, (6)

2.6 Example: Convection 171

w(0, t) = 0, hw(a, t) + κ∂w

∂x(a, t) = 0, 0 < t, (7)

w(x,0) = f (x) − v(x) ≡ g(x), 0 < x < a. (8)

The solution for w(x, t) can now be found by the product method. On theassumption that w has the form of a product φ(x)T(t), the variables can beseparated exactly as before, giving two ordinary differential equations linkedby a common parameter λ2:

φ′′ + λ2φ = 0, 0 < x < a,

T ′ + λ2kT = 0, 0 < t.

Also, since the boundary conditions are linear and homogeneous, they can betranslated directly into conditions on φ:

w(0, t) = φ(0)T(t) = 0,

κ∂w

∂x(a, t) + hw(a, t) = [

κφ′(a) + hφ(a)]T(t) = 0.

Either T(t) is identically zero (which would make w(x, t) identically zero) or

φ(0) = 0, κφ′(a) + hφ(a) = 0.

Combining the differential equation and boundary conditions on φ, we getthe eigenvalue problem

φ′′ + λ2φ = 0, 0 < x < a, (9)

φ(0) = 0, κφ′(a) + hφ(a) = 0. (10)



The boundary condition at x = 0 requires that φ(0) = c1 = 0, leaving φ(x) =c2 sin(λx). Now, at the other boundary,

κφ′(a) + hφ(a) = c2

(κλ cos(λa) + h sin(λa)

) = 0.

Discarding the possibilities c2 = 0 and λ = 0, which both lead to the trivialsolution, we are left with the equation

κλ cos(λa) + h sin(λa) = 0, or tan(λa) = −κ

hλ. (11)


A

n 0.2500 0.5000 1.0000 2.0000 4.00001 2.5704 2.2889 2.0288 1.8366 1.71552 5.3540 5.0870 4.9132 4.8158 4.76483 8.3029 8.0962 7.9787 7.9171 7.88574 11.3348 11.1727 11.0855 11.0408 11.01835 14.4080 14.2764 14.2074 14.1724 14.1548

Table 2 First five positive solutions of the equation tan(x) = −Ax

Figure 6 Graphs of tan(λa) and −λκ/h. The points of intersection are solutionsof tan(λa) = −λκ/h, eigenvalues of the problem Eqs. (9)–(10). The intersectionat λ = 0 corresponds to the trivial solution.

From sketches of the graphs of tan(λa) and −κλ/h (Fig. 6), we see that there isan infinite number of solutions, λ1, λ2, λ3, . . . , and that, for very large n, λn isgiven approximately by

λn∼= 2n − 1

2

π

a.

Table 2 shows the first five values of the product λa for several different valuesof the dimensionless parameter κ/ha. (More solutions are tabulated in Hand-book of Mathematical Functions by Abramowitz and Stegun.)

Thus we have for each n = 1,2, . . . an eigenvalue λ2n and an eigenfunction

φn(x), which satisfies the eigenvalue problem Eqs. (9) and (10). Accompanyingφn(x) is the function

Tn(t) = exp(−λ2

nkt)

2.6 Example: Convection 173

that makes wn(x, t) = φn(x)Tn(t) a solution of the partial differential equa-tion (6) and the boundary conditions Eq. (7). Since Eqs. (6) and (7) are lin-ear and homogeneous, any linear combination of solutions is also a solution.Therefore, the transient solution will have the form

w(x, t) =∞∑

n=1


nkt),

and the remaining condition to be satisfied, the initial condition Eq. (8), is

w(x,0) =∞∑

n=1

bn sin(λnx) = g(x), 0 < x < a. (12)

Thus the constants bn are to be chosen so as to make the infinite series equalg(x).

Although Eq. (12) looks like a Fourier series problem, it is not, because λ2,λ3, and so forth are not all integer multiples of λ1. If we attempt to use the ideaof orthogonality, we can still find a way to select the bn, for it may be shown bydirect computation that∫ a

0sin(λnx) sin(λmx)dx = 0, if n �= m. (13)

Then if we multiply both sides of the proposed Eq. (12) by sin(λmx) (wherem is fixed) and integrate from 0 to a, we have∫ a

0g(x) sin(λmx)dx =

∞∑n=1

bn

∫ a

0sin(λnx) sin(λmx)dx,

where we have integrated term by term. According to Eq. (13), all the termsof the series disappear except the one in which n = m, yielding an equationfor bm:

bm =∫ a

0 g(x) sin(λmx)dx∫ a0 sin2(λmx)dx

. (14)

By this formula, the bm may be calculated and inserted into the formula forw(x, t). Then we may put together the solution u(x, t) of the original problemEqs. (1)–(4):

u(x, t) = v(x) + w(x, t)

= T0 + xh(T1 − T0)

(κ + ha)+

∞∑n=1


nkt).

In Fig. 7 are graphs of u(x, t) for two different values of the parameter κ/ha;both have initial conditions u(x,0) = 0. See animations on the CD.


(a) (b)

Figure 7 Solution of Eqs. (1)–(4) with T0 = 20, T1 = 100, and f (x) = 0. Graphs(a) and (b) correspond to κ/ha = 0.1 and κ/ha = 1.0, respectively. In each case,u(x, t) is graphed as a function of x for times chosen so that the dimensionlesstime kt/a2 takes on the values 0.001, 0.01, 0.1, 1. Note that both the temperatureand its slope at the right end (x = a) change with time, so the boundary conditionEq. (3) is satisfied. See animations on the CD.

E X E R C I S E S

1. Sketch v(x) as given in Eq. (5) assuming

a. T1 > T0; b. T1 = T0; c. T1 < T0.

2. If T1 > T0, as in Fig. 7, what is the maximum value of the temperatureu(x, t) on the interval 0 ≤ x ≤ a at any fixed time t? The solution will be afunction of T0, T1 and z = κ/ha.

3. Why have we ignored the negative solutions of the equation

tan(λa) = −κλ

h?

4. Derive the formula Eq. (12) for the coefficients bm.

5. Sketch the first two eigenfunctions of this example taking κ/h = 0.5 (λ1 =2.29/a, λ2 = 5.09/a).

6. Verify that ∫ a

0sin2(λmx)dx = a

2+ κ

h

cos2(λma)

2.

2.7 Sturm–Liouville Problems 175

7. Find the coefficients bm corresponding to

g(x) = 1, 0 < x < a.

8. Using the solution of Exercise 7, write out the first few terms of the solu-tion of Eqs. (6)–(8), where g(x) = T, 0 < x < a.

9. Same as Exercise 7 for g(x) = x, 0 < x < a.

10. Verify the orthogonality integral by direct integration. It will be necessaryto use the equation that defines the λn:

κλn cos(λna) + h sin(λna) = 0.

2.7 Sturm–Liouville ProblemsAt the end of the preceding section, we saw that ordinary Fourier series arenot quite adequate for all the problems we can solve. We can make some gen-eralizations, however, that do cover most cases that arise from separation ofvariables. In simple problems, we often find eigenvalue problems of the form

φ′′ + λ2φ = 0, l < x < r, (1)

α1φ(l) − α2φ′(l) = 0, (2)

β1φ(r) + β2φ′(r) = 0. (3)

It is not difficult to determine the eigenvalues of this problem and to show theeigenfunctions orthogonal by direct calculation, but an indirect calculation isstill easier.

Suppose that φn and φm are eigenfunctions corresponding to differenteigenvalues λ2

n and λ2m. That is,

φ′′n + λ2

nφn = 0, φ′′m + λ2

mφm = 0,

and both functions satisfy the boundary conditions. Let us multiply the firstdifferential equation by φm and the second by φn, subtract the two, and movethe terms containing φnφm to the other side:

φ′′nφm − φ′′

mφn = (λ2

m − λ2n

)φnφm.

The right-hand side is a constant (nonzero) multiple of the integrand in theorthogonality relation∫ r

lφn(x)φm(x)dx = 0, n �= m,


which is proved true if the left-hand side is zero:∫ r

l

(φ′′

nφm − φ′′mφn

)dx = 0.

This integral is integrable by parts:∫ r

l

(φ′′

nφm − φ′′mφn

)dx

= [φ′

n(x)φm(x) − φ′m(x)φn(x)

]∣∣r

l−

∫ r

l

(φ′

nφ′m − φ′

mφ′n

)dx.

The last integral is obviously zero, so we have

(λ2

m − λ2n

)∫ r

lφn(x)φm(x)dx = [

φ′n(x)φm(x) − φ′

m(x)φn(x)]∣∣r

l.

Both φn and φm satisfy the boundary condition at x = r,

β1φm(r) + β2φ′m(r) = 0,

β1φn(r) + β2φ′n(r) = 0.

These two equations may be considered simultaneous equations in β1 and β2.At least one of the numbers β1 and β2 is different from zero; otherwise, therewould be no boundary condition. Hence the determinant of the equationsmust be zero:

φm(r)φ′n(r) − φn(r)φ′

m(r) = 0.

A similar result holds at x = l. Thus[φ′

n(x)φm(x) − φ′m(x)φn(x)

]∣∣r

l= 0,

and, therefore, we have proved the orthogonality relation∫ r

lφn(x)φm(x)dx = 0, n �= m,

for the eigenfunctions of Eqs. (1)–(3).We may make a much broader generalization about orthogonality of eigen-

functions with very little trouble. Consider the following model eigenvalueproblem, which might arise from separation of variables in a heat conductionproblem (see Section 9):[

s(x)φ′(x)]′ − q(x)φ(x) + λ2p(x)φ(x) = 0, l < x < r,

α1φ(l) − α2φ′(l) = 0,

β1φ(r) + β2φ′(r) = 0.


Let us carry out the procedure used in the preceding with this problem. Theeigenfunctions satisfy the differential equations(

sφ′n

)′ − qφn + λ2npφn = 0,(

sφ′m

)′ − qφm + λ2mpφm = 0.

Multiply the first by φm and the second by φn, subtract (the terms containingq(x) cancel), and move the term containing pφnφm to the other side:(

sφ′n

)′φm − (

sφ′m

)′φn = (

λ2m − λ2

n

)pφnφm. (4)

Integrate both sides from l to r, and apply integration by parts to the left-handside: ∫ r

l

[(sφ′

n

)′φm − (

sφ′m

)′φn

]dx

= [sφ′

nφm − sφ′mφn

]∣∣r

l−

∫ r

l

(sφ′

nφ′m − sφ′

nφ′m

)dx.

The second integral is zero. From the boundary conditions we find that

φ′n(r)φm(r) − φ′

m(r)φn(r) = 0,

φ′n(l)φm(l) − φ′

m(l)φn(l) = 0

by the same reasoning as before. Hence, we discover the orthogonality relation∫ r

lp(x)φn(x)φm(x)dx = 0, λ2

n �= λ2m

for the eigenfunctions of the problem stated.During these operations, we have made some tacit assumptions about in-

tegrability of functions after Eq. (4). In individual cases, where the coefficientfunctions s, q, and p and the eigenfunctions themselves are known, one caneasily check the validity of the steps taken. In general, however, we would liketo guarantee the existence of eigenfunctions and the legitimacy of computa-tions after Eq. (4). To do so, we need the following.

Definition

The problem

(sφ′)′ − qφ + λ2pφ = 0, l < x < r, (5)

α1φ(l) − α2φ′(l) = 0, (6)

β1φ(r) + β2φ′(r) = 0 (7)


is called a regular Sturm–Liouville problem if the following conditions are ful-filled:

a. s(x), s′(x), q(x), and p(x) are continuous for l ≤ x ≤ r;

b. s(x) > 0 and p(x) > 0 for l ≤ x ≤ r;

c. The α’s and β ’s are nonnegative, and α21 + α2

2 > 0, β21 + β2

2 > 0.

d. The parameter λ occurs only where shown.

Condition a and the first condition b guarantee that the differential equationhas solutions with continuous first and second derivatives. Notice that s(l) ands(r) must both be positive (not zero). Condition c just says that there are twoboundary conditions: α2

1 + α22 = 0 only if α1 = α2 = 0, which would be no

condition. The other requirements contribute to the desired properties in waysthat are not obvious.

We are now ready to state the theorems that contain necessary informationabout eigenfunctions.

Theorem 1. The regular Sturm–Liouville problem has an infinite number ofeigenfunctions φ1, φ2, . . . , each corresponding to a different eigenvalue λ2

1, λ22, . . . .

If n �= m, the eigenfunctions φn and φm are orthogonal with weight function p(x):∫ r

lφn(x)φm(x)p(x)dx = 0, n �= m. �

The theorem is already proved, for the continuity of coefficients and eigen-functions makes our previous calculations legitimate. It should be noted thatany constant multiple of an eigenfunction is also an eigenfunction; but asidefrom a constant multiplier, the eigenfunctions of a Sturm–Liouville problemare unique.

A number of other properties of the Sturm–Liouville problem are known.We summarize a few here.

Theorem 2. (a) The regular Sturm–Liouville problem has an infinite number ofeigenvalues, and λ2

n → ∞ as n → ∞.(b) If the eigenvalues are numbered in order, λ2

1 < λ22 < · · · , then the eigen-

function corresponding to λ2n has exactly n − 1 zeros in the interval l < x < r

(endpoints excluded).(c) If q(x) ≥ 0 and α1, α2, β1, β2, are all greater than or equal to zero, then all

the eigenvalues are nonnegative. �

Examples.

1. We note that the eigenvalue problems in Sections 3–6 of this chapter areall regular Sturm–Liouville problems, as is the problem in Eqs. (1)–(3) of


this section. In particular, the problem

φ′′ + λ2φ = 0, 0 < x < a,

φ(0) = 0, hφ(a) + κφ′(a) = 0

is a regular Sturm–Liouville problem, in which

s(x) = p(x) = 1, q(x) = 0, α1 = 1, α2 = 0,

β1 = h, β2 = κ.

All conditions of the definition are met.

2. A less trivial example is

(xφ′)′ + λ2

(1

x

)φ = 0, 1 < x < 2, φ(1) = 0, φ(2) = 0.

We identify s(x) = x, p(x) = 1/x, q(x) = 0. This is a regular Sturm–Liouville problem. The orthogonality relation is∫ 2

1φn(x)φm(x)

1

xdx = 0, n �= m.

The conclusions of Theorems 1 and 2 hold for both examples. �

E X E R C I S E S

1. The general solution of the differential equation in Example 2 is

φ(x) = c1 cos(λ ln(x)

) + c2 sin(λ ln(x)

).

Find the eigenvalues and eigenfunctions, and verify the orthogonality rela-tion directly by integration.

2. Check the results of Theorem 2 for the problem consisting of

φ′′ + λ2φ = 0, 0 < x < a,

with boundary conditions

a. φ(0) = 0, φ(a) = 0; b. φ′(0) = 0, φ′(a) = 0.

In case b, λ21 = 0.

3. Find the eigenvalues and eigenfunctions, and sketch the first few eigenfunc-tions of the problem

φ′′ + λ2φ = 0, 0 < x < a,


with boundary conditions

a. φ(0) = 0, φ′(a) = 0,

b. φ′(0) = 0, φ(a) = 0,

c. φ(0) = 0, φ(a) + φ′(a) = 0,

d. φ(0) − φ′(0) = 0, φ′(a) = 0,

e. φ(0) − φ′(0) = 0, φ(a) + φ′(a) = 0.

4. In Eqs. (1)–(3), take l = 0, r = a, and show that

a. The eigenfunctions are φn(x) = α2λn cos(λnx) + α1 sin(λnx).

b. The eigenvalues must be solutions of the equation

− tan(λa) = λ(α1β2 + α2β1)

α1β1 − α2β2λ2.

5. Show by applying Theorem 1 that the eigenfunctions of each of the follow-ing problems are orthogonal, and state the orthogonality relation.

a. φ′′ + λ2(1 + x)φ = 0, φ(0) = 0, φ′(a) = 0;

b. (exφ′)′ + λ2exφ = 0, φ(0) − φ′(0) = 0, φ(a) = 0;

c. φ′′ +(

λ2

x2

)φ = 0, φ(1) = 0, φ′(2) = 0;

d. φ′′ − sin(x)φ + exλ2φ = 0, φ′(0) = 0, φ′(a) = 0.

6. Consider the problem

(sφ′)′ − qφ + λ2pφ = 0, l < x < r,

φ(r) = 0,

in which s(l) = 0, s(x) > 0 for l < x ≤ r, but p and q satisfy the conditions ofa regular Sturm–Liouville problem. Require also that both φ(x) and φ′(x)have finite limits as x → l+. Show that the eigenfunctions (if they exist) areorthogonal.

7. The following problem is not a regular Sturm–Liouville problem. Why?Solve, and show that the eigenfunctions are not orthogonal.

φ′′ + λ2φ = 0, 0 < x < a,

φ(0) = 0, φ′(a) − λ2φ(a) = 0.

2.8 Expansion in Series of Eigenfunctions 181

8. Show that 0 is an eigenvalue of the problem

(sφ′)′ + λ2pφ = 0, l < x < r,

φ′(l) = 0, φ′(r) = 0,

where s and p satisfy the conditions of a regular Sturm–Liouville problem.

9. Find all values of the parameter µ for which there is a nonzero solution ofthis problem:

φ′′ + µφ = 0,

φ(0) + φ′(0) = 0, φ(a) + φ′(a) = 0.

One solution is negative. Does this contradict Theorem 2?

2.8 Expansion in Series of EigenfunctionsWe have seen that the eigenfunctions that arise from a regular Sturm–Liouvilleproblem

(sφ′)′ − qφ + λ2pφ = 0, l < x < r, (1)

α1φ(l) − α2φ′(l) = 0, (2)

β1φ(r) + β2φ′(r) = 0 (3)

are orthogonal with weight function p(x):∫ r

lp(x)φn(x)φm(x)dx = 0, n �= m, (4)

and it should be clear, from the way in which the question of orthogonalityarose, that we are interested in expressing functions in terms of eigenfunctionseries.

Suppose that a function f (x) is given in the interval l < x < r and that wewish to express f (x) in terms of the eigenfunctions φn(x) of Eqs. (1)–(3). Thatis, we wish to have

f (x) =∞∑

n=1

cnφn(x), l < x < r. (5)

The orthogonality relation Eq. (4) clearly tells us how to compute the coef-ficients. Multiplying both sides of the proposed Eq. (5) by φm(x)p(x) (where


m is a fixed integer) and integrating from l to r yields

∫ r

lf (x)φm(x)p(x)dx =

∞∑n=1

cn

∫ r

lφn(x)φm(x)p(x)dx.

The orthogonality relation says that all the terms in the series, except that onein which n = m, must disappear. Thus∫ r

lf (x)φm(x)p(x)dx = cm

∫ r

lφ2

m(x)p(x)dx

gives a formula for choosing cm.We can now cite a convergence theorem for expansion in terms of eigen-

functions. Notice the similarity to the Fourier series convergence theorem. Ofcourse, the Fourier sine or cosine series are series of eigenfunctions on a regu-lar Sturm–Liouville problem in which the weight function p(x) is 1.

Theorem. Let φ1, φ2, . . . be eigenfunctions of a regular Sturm–Liouville problemEqs. (1)–(3), in which the α’s and β ’s are not negative.

If f (x) is sectionally smooth on the interval l < x < r, then

∞∑n=1

cnφn(x) = f (x+) + f (x−)

2, l < x < r, (6)

where

cn =∫ r

l f (x)φn(x)p(x)dx∫ rl φ2

n(x)p(x)dx.

Furthermore, if the series

∞∑n=1

|cn|[∫ r

lφ2

n(x)p(x)dx

]1/2

converges, then the series Eq. (6) converges uniformly, l ≤ x ≤ r. �

E X E R C I S E S

1. Verify that

λ2n =

(nπ

ln(b)

)2

, φn = sin(λn ln(x)

)are the eigenvalues and eigenfunctions of

2.8 Expansion in Series of Eigenfunctions 183

(xφ′)′ + λ2

(1

x

)φ = 0, 1 < x < b,

φ(1) = 0, φ(b) = 0.

Find the expansion of the function f (x) = x in terms of these eigenfunc-tions. To what values does the series converge at x = 1 and x = b?

2. If φ1, φ2, . . . are the eigenfunctions of a regular Sturm–Liouville problemand are orthogonal with weight function p(x) on l < x < r and if f (x) issectionally smooth, then

∫ r

lf 2(x)p(x)dx =

∞∑n=1

anc2n,

where

an =∫ r

lφ2

n(x)p(x)dx

and cn is the coefficient of f as given in the theorem. Show why this shouldbe true, and conclude that cn

√an → 0 as n → ∞.

3. Verify that the eigenvalues and eigenfunctions of the problem

(exφ′)′ + exγ 2φ = 0, 0 < x < a,

φ(0) = 0, φ(a) = 0

are

γ 2n =

(nπ

a

)2

+ 1

4, φn(x) = exp

(−x

2

)sin

(nπx

a

).

Find the coefficients for the expansion of the function f (x) = 1, 0 < x < a,in terms of the φn.

4. If φ1, φ2, . . . are eigenfunctions of a regular Sturm–Liouville problem, thenumbers

√an are called normalizing constants, and the functions ψn =

φn/√

an are called normalized eigenfunctions. Show that∫ r

lψ2

n(x)p(x)dx = 1,

∫ r

lψn(x)ψm(x)p(x)dx = 0, n �= m.

5. Find the formula for the coefficients of a sectionally smooth function f (x)in the series

f (x) =∞∑

n=1

bnψn(x), l < x < r,

where the ψn are normalized eigenfunctions.


6. Show that, for the function in Exercise 5,

∫ r

lf 2(x)p(x)dx =

∞∑n=1

b2n.

7. What are the normalized eigenfunctions of the following problem?

φ′′ + λ2φ = 0, 0 < x < 1,

φ′(0) = 0, φ′(1) = 0.

2.9 Generalities on the Heat Conduction ProblemOn the basis of the information we have about the Sturm–Liouville problem,we can make some observations on a fairly general heat conduction problem.We take as a physical model a rod whose lateral surface is insulated. In orderto simplify slightly, we will assume that no heat is generated inside the rod.

Since material properties may vary with position, the partial differentialequation that governs the temperature u(x, t) in the rod will be

∂

∂x

(κ(x)

∂u

∂x

)= ρ(x)c(x)

∂u

∂t, l < x < r, 0 < t. (1)

Any of the three types of boundary conditions may be imposed at eitherboundary, so we use as boundary conditions

α1u(l, t) − α2∂u

∂x(l, t) = c1, t > 0, (2)

β1u(r, t) + β2∂u

∂x(r, t) = c2, t > 0. (3)

If the temperature is fixed, the coefficient of ∂u/∂x is zero. If the boundary isinsulated, the coefficient of u is zero, and the right-hand side is also zero. Ifthere is convection at a boundary, both coefficients will be positive, and thesigns will be as shown.

We already know that in the case of two insulated boundaries, the steady-state solution has some peculiarities, so we set this aside as a special case. As-sume, then, that either α1 or β1 or both are positive. Finally we need an initialcondition in the form

u(x,0) = f (x), l < x < r. (4)

Equations (1)–(4) make up an initial value–boundary value problem.

2.9 Generalities on the Heat Conduction Problem 185

Assuming that c1 and c2 are constants, we must first find the steady-statesolution

v(x) = limt→∞ u(x, t).

The function v(x) satisfies the boundary value problem

d

dx

(κ(x)

dv

dx

)= 0, l < x < r, (5)

α1v(l) − α2v′(l) = c1, (6)

β1v(r) + β2v′(r) = c2. (7)

Since we have assumed that at least one of α1 or β1 is positive, this problemcan be solved. In fact, it is possible to give a formula for v(x) in terms of thefunction (see Exercise 1) ∫ x

l

dξ

κ(ξ)= I(x). (8)

Before proceeding further, it is convenient to introduce some new functions.Let κ , ρ, and c indicate average values of the functions κ(x), ρ(x), and c(x).We shall define dimensionless functions s(x) and p(x) by

κ(x) = κs(x), ρ(x)c(x) = ρ cp(x).

Also, we define the transient temperature to be

w(x, t) = u(x, t) − v(x).

By direct computation, using the fact that v(x) is a solution of Eqs. (5)–(7),we can show that w(x, t) satisfies the initial value–boundary value problem

∂

∂x

(s(x)

∂w

∂x

)= 1

kp(x)

∂w

∂t, l < x < r, 0 < t, (9)

α1w(l, t) − α2∂w

∂x(l, t) = 0, 0 < t, (10)

β1w(r, t) + β2∂w

∂x(r, t) = 0, 0 < t, (11)

w(x,0) = f (x) − v(x) = g(x), l < x < r, (12)

which has homogeneous boundary conditions. The constant k is defined to beκ/ρ c.

Now we use our method of separation of variables to find w. If w has theform w(x, t) = φ(x)T(t), the differential equation becomes

T(t)(s(x)φ′(x)

)′ = 1

kp(x)φ(x)T ′(t),


and, on dividing through by pφT, we find the separated equation

(sφ′)′

pφ= T ′

kT, l < x < r, 0 < t.

As before, the equality between a function of x and a function of t can holdonly if their common value is constant. Furthermore, we expect the constantto be negative, so we put

(sφ′)′

pφ= T ′

kT= −λ2

and separate two ordinary equations

T ′ + λ2kT = 0, 0 < t,

(sφ′)′ + λ2pφ = 0, l < x < r.

The boundary conditions, being linear and homogeneous, can also be changedinto conditions of φ. For instance, Eq. (10) becomes[

α1φ(l) − α2φ′(l)

]T(t) = 0, 0 < t,

and, because T(t) ≡ 0 makes w(x, t) ≡ 0, we take the other factor to be zero.We have, then, the eigenvalue problem

(sφ′)′ + λ2pφ = 0, l < x < r, (13)

α1φ(l) − α2φ′(l) = 0, (14)

β1φ(r) + β2φ′(r) = 0. (15)

Since s and p are related to the physical properties of the rod, they should bepositive. We suppose also that s, s′, and p are continuous. Then Eqs. (13)–(15)comprise a regular Sturm–Liouville problem, and we know the following.

1. There is an infinite number of eigenvalues

0 < λ21 < λ2

2 < · · · .

2. To each eigenvalue corresponds just one eigenfunction (give or take a con-stant multiplier).

3. The eigenfunctions are orthogonal with weight p(x):∫ r

lφn(x)φm(x)p(x)dx = 0, n �= m.

The function Tn(t) that accompanies φn(x) is given by

Tn(t) = exp(−λ2

nkt).

2.9 Generalities on the Heat Conduction Problem 187

We now begin to assemble the solution. For each n = 1,2,3, . . . ,wn(x, t)= φn(x)Tn(t) satisfies Eqs. (9)–(11). As these are all linear homogeneous equa-tions, any linear combination of solutions is again a solution. Thus the tran-sient temperature has the form

w(x, t) =∞∑

n=1

anφn(x) exp(−λ2

nkt).

The initial condition Eq. (12) will be satisfied if we choose the an so that

w(x,0) =∞∑

n=1

anφn(x) = g(x), l < x < r.

The convergence theorem tells us that the equality will hold, except possibly ata finite number of points, if f (x) — and therefore g(x) — is sectionally smooth.Thus w(x, t) is the solution of its problem, if we choose

an =∫ r

l g(x)φn(x)p(x)dx∫ rl φ2

n(x)p(x)dx.

Finally, we can write the complete solution of Eqs. (1)–(4) in the form

u(x, t) = v(x) +∞∑

n=1

anφn(x) exp(−λ2

nkt). (16)

Working from the representation Eq. (16) we can draw some conclusionsabout the solution of Eqs. (1)–(4).

1. Since all the λ2n are positive, u(x, t) does tend to v(x) as t → ∞.

2. For any t1 > 0, the series for u(x, t1) converges uniformly in l ≤ x ≤ rbecause of the exponential factors; therefore u(x, t1) is a continuous func-tion of x. Any discontinuity in the initial condition is immediately elimi-nated.

3. For large enough values of t, we can approximate u(x, t) by

v(x) + a1φ1(x) exp(−λ2

1kt).

(To judge how large t might be, we need to know something about the an

and the λn.) Because φ1(x) is of one sign on the interval l < x < r (thatis, φ1(x) > 0 or φ1(x) < 0 for all x between l and r), the graph of ourapproximation will lie either above or below the graph of v(x) but willnot cross it (provided that a1 �= 0).


E X E R C I S E S

1. Find the explicit form for v(x) in terms of the function in Eq. (8) assuming

a. α1 = β1 = 0, c1 = c2 = 0;

b. α1 > 0 or β1 > 0, and no coefficient negative.

Why are these two cases separate?

2. Justify each of the conclusions.

3. Derive the general form of u(x, t) if the boundary conditions are ∂u/∂x = 0at both ends. In this case, λ2 = 0 is an eigenvalue.

2.10 Semi-Infinite RodUp to this point we have seen only problems over finite intervals. Frequently,however, it is justifiable and useful to assume that an object is infinite in length.(Sometimes this assumption is used to disguise ignorance of a boundary con-dition or to suppress the influence of a complicated condition.) Thus, if the rodwe have been studying is very long, we may treat it as semi-infinite — that is, asextending from 0 to ∞. If properties are uniform and there is no “generation,”the partial differential equation governing the temperature u(x, t) remains

∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t.

Let us suppose that at x = 0 the temperature is held constant, say, u(0, t) = 0in some temperature scale. In the absence of another boundary, there is noother boundary condition. However, it is desirable that u(x, t) remain finite —less than some fixed bound — as x → ∞.

Thus, our mathematical model is

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < ∞, 0 < t, (1)

u(0, t) = 0, 0 < t, (2)

u(x, t) bounded as x → ∞, (3)

u(x,0) = f (x), 0 < x. (4)

The heat equation (1) and the boundary condition (2) are homogeneous.The boundedness condition (3) is also homogeneous in an important way:A (finite) sum of bounded functions is bounded. Thus, we can attack Eqs. (1)–(3) by separation of variables. Assume that u(x, t) = φ(x)T(t), so the partial

2.10 Semi-Infinite Rod 189

differential equation can be separated into two ordinary equations as usual:

φ′′(x)

φ(x)= T ′(t)

kT(t)= const. (5)

There is just one boundary condition on u, which requires that φ(0) = 0. Theboundedness condition also requires that φ(x) remain finite as x → ∞. It iseasy to check (see Exercises) that a positive separation constant produces func-tions φ(x) that cannot fulfill both the boundary and boundedness conditionswithout being identically 0. Thus, we must choose a negative separation con-stant, −λ2. The differential equation, together with the boundary and bound-edness conditions, forms a singular eigenvalue problem (singular because ofthe semi-infinite interval),

φ′′ + λ2φ = 0, 0 < x, (6)

φ(0) = 0, φ(x) bounded as x → ∞. (7)


φ(x) = c1 cos(λx) + c2 sin(λx),

which is bounded for any choice of the constants and for any value of λ. Theboundedness condition told us to use a negative constant in Eq. (5) and nowcontributes nothing further.

Applying the boundary condition at x = 0 shows that c1 = 0, leaving φ(x) =c2 sin(λx). In this singular eigenvalue problem, there are no “special” valuesof λ: Any value produces a nonzero solution of the differential equation thatalso satisfies the boundary and boundedness conditions. (But negative valuesof λ produce no new solutions.) Recalling that any constant multiple of a so-lution of a homogeneous problem is still a solution, we choose c2 = 1 andsummarize the solution of the singular eigenvalue problem as

φ(x;λ) = sin(λx), λ > 0. (8)

The solution of Eq. (5) for T(t), with constant −λ2, is

T(t) = exp(−λ2kt

).

For any value of λ2, the function

u(x, t;λ) = sin(λx) exp(−λ2kt

)satisfies Eqs. (1)–(3). Equation (1) and the boundary condition Eq. (2) are ho-mogeneous, and Eq. (3) is homogeneous in effect; therefore any linear combi-nation of solutions is a solution. Since the parameter λ may take on any value,


we must use an integral — the continuous analogue of a sum or series — toinclude all possibilities. Thus u should have the form

u(x, t) =∫ ∞

0B(λ) sin(λx) exp

(−λ2kt)dλ. (9)

(We need not include negative values of λ. They give no new solutions.) Theinitial condition will be satisfied if B(λ) is chosen to make

u(x,0) =∫ ∞

0B(λ) sin(λx)dλ = f (x), 0 < x.

We recognize this as a Fourier integral; B(λ) is to be chosen as

B(λ) = 2

π

∫ ∞

0f (x) sin(λx)dx. (10)

If B(λ) exists, then Eq. (9) is the solution of the problem. Notice that whent > 0, the exponential function makes the improper integral in Eq. (9) con-verge very rapidly.

Some care must be taken in the interpretation of our solution. If the rod re-ally is finite (say, length L) the expression in Eq. (9) is, of course, meaninglessfor x greater than L. The presence of a boundary condition at x = L wouldinfluence temperatures nearby, so Eq. (9) can be considered a valid approxi-mation only for x L.

Example.Solve the problem in Eqs. (1)–(4) using the initial temperature distribution

f (x) ={

T0, 0 < x < b,0, b < x.

This means that a section of length b at the left end of the rod starts out attemperature T0, different from the temperature of the long right end, whichis at the same temperature as the left boundary. (We assume T0 > 0.) Thesolution is given by Eq. (9), with B(λ) calculated from Eq. (10):

B(λ) = 2

π

∫ ∞

0f (x) sin(λx)dx

= 2

π

∫ b

0T0 sin(λx)dx

= 2T0

λπ

(1 − cos(λb)

).

2.10 Semi-Infinite Rod 191

Figure 8 Graphs of the solution of the example, u(x, t) as a function of x overthe interval 0 < x < 3b, where b = 1 and T0 = 100 for convenience. The timeshave been chosen so that the dimensionless time kt/b2 takes the values 0.001, 0.01,0.1, and 1. When kt/b2 = 0.01, the temperature near x = b/2 has not changednoticeably from its initial value.

Therefore, the complete solution is

u(x, t) = 2

πT0

∫ ∞

0

1 − cos(λb)

λsin(λx) exp

(−λ2kt)

dλ.

In Fig. 8 are graphs of u(x, t) as a function of x for various values of t; ananimation can be seen on the CD. �

E X E R C I S E S

1. Find the solution of Eqs. (1)–(3) if the initial temperature distribution isgiven by

f (x) ={0, 0 < x < a,

T, a < x < b,0, b < x.

2. Verify that u(x, t) as given by Eq. (9) is a solution of Eqs. (1)–(3). What isthe steady-state temperature distribution?

3. Find the solution of Eqs. (1)–(4) if f (x) = T0e−αx, x > 0.

4. Find a formula for the solution of the problem


∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t,

∂u

∂x(0, t) = 0, 0 < t,

u(x,0) = f (x), 0 < x.

5. Determine the solution of Exercise 4 if f (x) is the function given in Exer-cise 1.

6. Penetration of heat into the earth. Assume that the earth is flat, occupyingthe region 0 < x (so that x measures distance down from the surface). Atthe surface, the temperature fluctuates according to season, time of day, etc.We cover several cases by taking the boundary condition to be u(0, t) =sin(ωt), where the frequency ω can be chosen according to the period ofinterest.

a. Show that u(x, t) = e−px sin(ωt − px) satisfies the boundary conditionand is a solution of the heat equation if p = √

ω/2k.

b. Sketch u(x, t) as a function of t for x = 0,1, and 2 m, taking ω = 2 ×10−7 rad/s (approximately one cycle per year) and k = 0.5 × 10−6 m2/s.

c. With ω as in part b, find the depth (as a function of k) at which seasonsare reversed.

7. Consider the problem

∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t,

u(0, t) = T0, 0 < t,

u(x,0) = f (x), 0 < x.

Show that, for our method of solution to work, it is necessary to have T0 =limx→∞ f (x). Find a formula for u(x, t) if this is the case.

8. If the separation constant in Eq. (5) were positive (say, p2), we would at-tempt to solve φ′′ − p2φ = 0 subject to the conditions, Eq. (7). Solve thedifferential equation, and show that any nonzero, bounded solution is not0 at x = 0 and that any solution that is 0 at x = 0 is not bounded.

9. R.C. Bales, M.P. Valdez and G.A. Dawson [Gaseous deposition to snow,2: Physical-chemical model for SO2 deposition, Journal of Geophysical Re-search, 92 (1987): 9789–9799] develop a mathematical model for the trans-port of SO2 gas into snow by molecular diffusion. The governing partial

2.11 Infinite Rod 193

differential equation is

∂C

∂t= D

(∂2C

∂x2− a2C

),

where C is the concentration of SO2 as a function of x (depth into the snow)and time and D is a diffusion constant. The term containing C appears be-cause the SO2 takes part in a chemical reaction with water in the snow,forming sulphuric acid, H2SO4. The coefficient a2 depends on pH, temper-ature, and other circumstances; we treat it as a constant. The problem is tobe solved for a wide range of values for the parameters.

If the snow is deep, the authors believe that it is reasonable to use a semi-infinite interval for x and to add the condition C(x, t) → 0 as x → ∞. Inaddition, a natural boundary condition at the snow surface is that con-centration in the snow match that in the air: C(0, t) = C0. Furthermore,if the snow is fresh, we can assume that the concentration throughout isinitially 0,

C(x,0) = 0, 0 < x.

a. Find a steady-state solution v(x) that satisfies the partial differentialequation and the boundary conditions.

b. State the problem (partial differential equation, boundary condition atx = 0, condition as x → ∞, and initial condition) to be satisfied by thetransient w(x, t) = C(x, t) − v(x).

c. Solve the problem for the transient. Note that the condition as x → ∞must be relaxed to: w(x, t) bounded as x → ∞. Individual product so-lutions do not approach 0 as x increases.

2.11 Infinite Rod

If we wish to study heat conduction in the center of a very long rod, we may as-sume that it extends from −∞ to ∞. Then there are no boundary conditions,and the problem to be solved is

∂2u

∂x2= 1

k

∂u

∂t, −∞ < x < ∞, 0 < t, (1)

u(x,0) = f (x), −∞ < x < ∞, (2)∣∣u(x, t)∣∣ bounded as x → ±∞. (3)


Using the same techniques as before, we look for solutions in the formu(x, t) = φ(x)T(t) so that the heat equation (1) becomes

φ′′(x)

φ(x)= T ′(t)

T(t)= constant.

As in the previous section, the constant must be nonpositive (say, −λ2) inorder for the solutions to be bounded. Thus, we have the singular eigenvalueproblem

φ′′ + λ2φ = 0, −∞ < x < ∞,

φ(x) bounded as x → ±∞.

It is easy to see that every solution of φ′′/φ = −λ2 is bounded. Thus, our fac-tors φ(x) and T(t) are

φ(x;λ) = A cos(λx) + B sin(λx),

T(t;λ) = exp(−λ2kt

).

We combine the solutions φ(x)T(t) in the form of an integral to obtain

u(x, t) =∫ ∞

0


)exp

(−λ2kt)

dλ. (4)

At time t = 0, the exponential factor becomes 1, and the initial condition is∫ ∞

0


)dλ = f (x), −∞ < x < ∞.

As this is clearly a Fourier integral problem, we must choose A(λ) and B(λ) tobe the Fourier integral coefficient functions,

A(λ) = 1

π

∫ ∞

−∞f (x) cos(λx)dx, B(λ) = 1

π

∫ ∞

−∞f (x) sin(λx)dx. (5)

Then the function u(x, t) in Eq. (4) satisfies the partial differential equation (1)and the initial condition (2), provided that f is sectionally smooth and |f (x)|has a finite integral. It can be proved that the boundedness condition (3) isalso satisfied, provided that the initial value f (x) is bounded as x → ±∞.

Example.Solve the problem posed in Eqs. (1)–(3) with

f (x) ={0, x < −a,

T0, −a < x < a,0, a < x.


Figure 9 Solution of example problem. At t = 0, the temperature is T0 > 0 for−a < x < a and is 0 in the rest of the rod; u(x, t) is shown as a function of xon the interval −3a < x < 3a for three times. The times are chosen so that thedimensionless time kt/a2 takes the values 0.01, 1, and 10 (to get a clear picture ofthe changes in u). Note that u(x, t) is positive everywhere for any t > 0. The valuesT0 = 100 and a = 1 have been used for convenience. Also see the CD.

In words, the rod has a center section of length 2a whose temperature is differ-ent from that of the long sections to the left and right. We must compute thecoefficient functions A(λ) and B(λ). The latter is identically 0 because f (x) isan even function; and

A(λ) = 1

π

∫ ∞

−∞f (x) cos(λx)dx

= 1

π

∫ a

−aT0 cos(λx)dx

= 2T0

λπsin(λa).

Thus, the solution of the problem is

u(x, t) = 2T0

π

∫ ∞

0

sin(λa)

λcos(λx) exp

(−λ2kt)

dλ. (6)

This function is graphed as a function of x for several values of t in Fig. 9and animated on the CD. The figure suggests that u(x, t) is positive for all xwhen t > 0. This is indeed true and illustrates an interesting property of thesolutions of the heat equation: the instantaneous transmission of information.The “hot” section in the interval −a < x < a instantly raises the temperatureeverywhere else from the initial value of 0 to a positive value. �

Starting from the general form of a solution in Eq. (4), we can derive somevery interesting results. Change the variable of integration in Eq. (5) to x′ and


substitute the formulas for A(λ) and B(λ) into Eq. (4):

u(x, t) = 1

π

∫ ∞

0

[∫ ∞

−∞f (x′) cos(λx′)dx′ cos(λx)

+∫ ∞

−∞f (x′) sin(λx′)dx′ sin(λx)

]exp

(−λ2kt)

dλ.

Combining terms we find

u(x, t) = 1

π

∫ ∞

0

∫ ∞

−∞f (x′)

[cos(λx′) cos(λx) + sin(λx′) sin(λx)

]dx′

× exp(−λ2kt

)dλ

= 1

π

∫ ∞

0

∫ ∞

−∞f (x′) cos

(λ(x′ − x)

)dx′ exp

(−λ2kt)

dλ.

If the order of integration may be reversed, we may write

u(x, t) = 1

π

∫ ∞

−∞f (x′)

∫ ∞

0cos

(λ(x′ − x)

)exp

(−λ2kt)

dλdx′.

The inner integral can be computed by complex methods of integration. Itis known to be (Miscellaneous Exercises 32, Chapter 1)

∫ ∞

0cos

(λ(x′ − x)

)exp

(−λ2kt)

dλ =√

π

4ktexp

[−(x′ − x)2

4kt

], t > 0.

This gives us, finally, a new form for the temperature distribution:

u(x, t) = 1√4kπ t

∫ ∞

−∞f (x′) exp

[−(x′ − x)2

4kt

]dx′. (7)

Using this form, we find the solution of the example problem solved earlier(see Eq. (6)) to be

u(x, t) = T0√4πkt

∫ a

−aexp

[−(x′ − x)2

4kt

]dx′. (8)

Of the two formulas, Eqs. (4) and (7), for the solution u(x, t), each has itsadvantages. For simple problems we may be able to evaluate the coefficientsA(λ) and B(λ) in Eq. (5). However, it is a rare case indeed when the integralin Eq. (4) can be evaluated analytically. The same is true for the integral inEq. (7). Thus, if the value of u at a specific x and t is needed, either integralwould be calculated numerically. For large values of kt, the exponential factor


in the integrand of Eq. (4) will be nearly zero, except for small λ. Thus, Eq. (4)is approximately

u(x, t) ∼=∫ �

0


)exp

(−λ2kt)

dλ

for � not large, and the right-hand side may be found to a high degree ofaccuracy with little effort.

On the other hand, if kt is small, the exponential in the integrand of Eq. (7)will be nearly zero, except for x′ near x. The approximation

u(x, t) ∼= 1√4kπ t

∫ x+h

x−hf (x′) exp

[−(x′ − x)2

4kt

]dx′

is satisfactory for h not large, and again numerical techniques are easily appliedto the right-hand side.

The expression in Eq. (7) also has a number of other advantages. It requiresno intermediate integrations (compare Eq. (5)). It shows directly the influenceof initial conditions on the solution. Moreover, the function f (x) need notsatisfy the restriction ∫ ∞

−∞

∣∣ f (x)∣∣dx < ∞

in order for Eq. (7) to satisfy the original problem.

E X E R C I S E S

See the exercise Common Singular Eigenvalue Problems on the CD.1. Find the solution of Eqs. (1)–(3) using the form given in Eq. (7) if the initial

temperature distribution is

f (x) ={

T0, x < 0,T1, 0 < x.

2. Find the solution of Eqs. (1)–(3) using the form given in Eq. (4) if

f (x) ={

T0

(a − |x|), −a < x < a,

0, otherwise.

3. Same task as in Exercise 2, with f (x) = T0e−|x/a| for all x.

4. Show that the solution of the problem studied in Section 10,

∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t,


u(0, t) = 0, 0 < t,

u(x,0) = f (x), 0 < x,

can be expressed as

u(x, t) = 1√4πkt

∫ ∞

0f (x′)

[exp

(−(x′ − x)2

4kt

)− exp

(−(x′ + x)2

4kt

)]dx′.

Hint: Start from the problem of this section with initial condition

u(x,0) = fo(x), −∞ < x < ∞,

where fo is the odd extension of f . Then use Eq. (7), and split the intervalof integration at 0.

5. Verify by differentiating that the function

u(x, t) = 1√4kπ t

exp

[− x2

4kt

]

is a solution of the heat equation

∂2u

∂x2= 1

k

∂u

∂t, 0 < t, −∞ < x < ∞.

What can be said about u at x = 0? at t = 0+? What is limt→0+ u(0, t)?Sketch u(x, t) for various fixed values of t.

6. Suppose that f (x) is an odd periodic function with period 2a. Show thatu(x, t) defined by Eq. (7) also has these properties.

7. If f (x) = 1 for all x, the solution of our heat conduction problem isu(x, t) = 1. Use this fact together with Eq. (7) to show that

1 = 1√4πkt

∫ ∞

−∞exp

[−(x′ − x)2

4kt

]dx′.

8. Solve the problem that follows using Eq. (7).

∂2u

∂x2= 1

k

∂u

∂t, −∞ < x < ∞, 0 < t,

u(x,0) ={1, x > 0,

−1, x < 0.

9. Can Exercise 8 be solved in the form of Eq. (4)? Note that

2

π

∫ ∞

0

sin(λx)

λdλ =

{1, 0 < x,−1, x < 0.

2.12 The Error Function 199

Figure 10 Graph of the error function erf(z) for −3 < z < 3.

2.12 The Error FunctionIn Section 11 we made transformations of a Fourier integral to obtain the so-lution of the heat problem

∂2u

∂x2= 1

k

∂u

∂t, −∞ < x < ∞, 0 < t, (1)

u(x,0) = f (x), −∞ < x < ∞, (2)

in the form of a single integral,

u(x, t) = 1√4πkt

∫ ∞

−∞f (x′)e−(x−x′)2/4kt dx′. (3)

Even for the simplest functions f , this integration cannot be carried out inclosed form, mainly because the indefinite integral

∫e−x2

dx is not an elemen-tary function. We can improve our understanding of the solution Eq. (3) if weintroduce the error function, defined as

erf(z) = 2√π

∫ z

0e−y2

dy. (4)

A graph of erf(z) is shown in Fig. 10. Convenient tables, together with approx-imations to the error function, will be found in Handbook of MathematicalFunctions, by Abramowitz and Stegun.

Several important properties of the error function follow immediately fromthe definition. First, it is clear that erf(0) = 0, and it is easy to show that erf isan odd function (Exercise 1). Second, by the fundamental theorem of calculus,the derivative of the error function is

d

dzerf(z) = 2√

πe−z2

. (5)


And finally, the error function supplies the integral

∫ b

ae−y2

dy =√

π

2

(erf(b) − erf(a)

). (6)

The reason for the choice of the constant in front of the integral in Eq. (4) isto make

limz→∞ erf(z) = 1. (7)

To see that this is true, define

A =∫ ∞

0e−y2

dy.

We are going to show that A = √π/2. First write A2 as the product of two

integrals,

A2 =∫ ∞

0e−y2

dy

∫ ∞

0e−x2

dx.

Remember that the name of the variable of integration in a definite integralis immaterial. This expression for A2 can be interpreted as an iterated doubleintegral over the first quadrant of the x,y-plane, equivalent to

A2 =∫ ∞

0

∫ ∞

0e−(x2+y2) dx dy.

Now change to polar coordinates. The first quadrant is described by the in-equalities 0 < r < ∞, 0 < θ < π/2, and the element of area in polar coordi-nates is r dr dθ . Thus, we have

A2 =∫ π/2

0

∫ ∞

0e−r2

r dr dθ. (8)

This integral, which can be evaluated by elementary means (see Exercise 2),has value π/4. Hence A = √

π/2 and Eq. (7) is validated.Many workers also use the complementary error function, erfc(z), defined as

erfc(z) = 2√π

∫ ∞

ze−y2

dy. (9)

By using Eq. (7) we obtain the identity

erfc(z) = 1 − erf(z). (10)

Some properties of the complementary error function are found in Exercise 3.


We are interested in the error function because of its role in solving the heatequation. First we shall show that the solution of the problem

∂2u

∂x2= 1

k

∂u

∂t, −∞ < x < ∞, 0 < t, (11)

u(x,0) = sgn(x), −∞ < x < ∞ (12)

is u(x, t) = erf(x/√

4kt). (Recall that sgn(x) has the value −1 if x is negative or+1 if x is positive.) The easy way to prove this statement is to verify it directly.(See Exercises 4 and 5.) Here, we shall arrive at the same conclusion, startingfrom Eq. (3),

u(x, t) = 1√4πkt

∫ ∞

−∞sgn(x′)e−(x−x′)2/4kt dx′. (13)

First, change the variable of integration to y = (x′ − x)/√

4kt. Then dy =dx′/

√4kt, and

u(x, t) = 1√π

∫ ∞

−∞sgn

(x + y

√4kt

)e−y2

dy. (14)

Now the function e−y2is even, and the sgn function changes from −1 to +1 at

y = −x/√

4kt. Thus, the integrand of Eq. (14) is as shown in Fig. 11. The tailto the left of −x/

√4kt has the same area as the tail to the right of x/

√4kt but

opposite sign. These two areas cancel, leaving

u(x, t) = 1√π

∫ x/√

4kt

−x/√

4kte−y2

dy.

Finally, use the symmetry of the integrand to halve the interval of integra-tion and double the result:

u(x, t) = 2√π

∫ x/√

4kt

0e−y2

dy = erf(x/

√4kt

).

This is the result we wanted to arrive at. Figure 12 shows graphs of u(x, t) =erf(x/

√4kt) as a function of x for several values of kt.

Because erf(0) = 0, the function u(x, t) = erf(x/√

4kt) must also be the so-lution of the problem

∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t,

u(0, t) = 0, 0 < t,

u(x,0) = 1, 0 < x.


(a)

(b)

Figure 11 (a) Graph of exp(−y2) and (b) graph of sgn(x + y√

4kt) exp(−y2).The tails beyond ±x/

√4kt have the same areas, with opposite signs.

Figure 12 Graphs of the solution of the problem in Eqs. (11) and (12),u(x, t) = erf(x/

√4kt), for x in the range −2 to 2 and for kt = 0.01, 0.1, 1, and 10.

As kt increases, the graph of u(x, t) collapses toward the x-axis.

A simple modification leads to the conclusion that the complementary errorfunction, u(x, t) = erfc(x/

√4kt) is the solution of this problem with zero ini-

tial condition and constant boundary condition,

∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t,

u(0, t) = 1, 0 < t,

u(x,0) = 0, 0 < x.


E X E R C I S E S

1. Show that erf(−z) = − erf(z), that is, that erf is an odd function.

2. Carry out the integration indicated in Eq. (8).

3. Verify these properties of the complementary error function:

a.d

dzerfc(z) = −e−z2 2√

π;

b. erfc(0) = 1;

c. limz→∞ erfc(z) = 0;

d. limz→−∞ erfc(z) = 2;

e. erfc(z) is neither even nor odd.

4. Verify by differentiating that u(x, t) = erf(x/√

4kt) satisfies the heat equa-tion (1).

5. Verify that u(x, t) = erf(x/√

4kt) satisfies the initial condition

u(x, t) ={1, 0 < x,

−1, x < 0.

6. In probability and statistics, the normal, or Gaussian, probability densityfunction is defined as

f (z) = 1√2π

e−z2/2, −∞ < z < ∞,

and the cumulative distribution function is

�(x) =∫ x

−∞f (z)dz.

Show that the cumulative distribution function and the error function arerelated by �(x) = [1 + erf(x/

√2)]/2.

7. Express this integral in terms of the error function:

I(x) =∫

e−x

√x

dx.

8. Use error functions to solve the problem

∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t,


u(0, t) = Ub, 0 < t,

u(x,0) = Ui, 0 < x.

(Hint: What conditions does u(x, t) − Ub satisfy?)

9. Assuming that Ub < 0 and Ui > 0, the problem in Exercise 8 might beinterpreted as representing the temperature in a freezing lake. (Think ofx as measuring depth from the surface.) Define x(t) as the depth of theice–water interface; then u(x(t), t) = 0. Now find x(t) explicitly.

10. Use error functions to solve the problem

∂2u

∂x2= 1

k

∂u

∂t, −∞ < x < ∞, 0 < t,

u(x,0) = f (x), −∞ < x < ∞,

where f (x) = U0 for x < 0 and f (x) = U1 for x > 0.

2.13 Comments and ReferencesIn about 1810, Fourier made an intensive study of heat conduction problems,in which he used the product method of solution and developed the idea ofFourier series. Sturm and Liouville made their clear and simple generalizationof Fourier series in the 1830s. Among modern works, Conduction of Heat inSolids, by Carslaw and Jaeger, is the standard reference. The Mathematics ofDiffusion, by Crank, and The Heat Equation, by D.V. Widder, are also usefulreferences. (See the Bibliography.)

Although we have motivated our study in terms of heat conduction and,to a lesser extent, by diffusion, many other physical phenomena of interest inengineering are described by the heat/diffusion equation: for example, voltageand current in an inductance-free cable and vorticity transport in fluid flow.The heat/diffusion equation and allied equations are being employed in biol-ogy to model cell physiology, chemical reactions, nerve impulses, the spread ofpopulations, and many other phenomena. Two good references are Differen-tial Equations and Mathematical Biology, by D.S. Jones and B.D. Sleeman, andMathematical Biology, by J.D. Murray.

The diffusion equation also turns up in some classical problems of probabil-ity theory, especially the description of Brownian motion. Suppose a particlemoves exactly one step of length x in each time interval t. The step maybe either to the left or to the right, each equally likely. Let ui(m) denote theprobability that, at time mt, the particle is at point ix (m = 0,1,2, . . . ,i = 0,±1,±2, . . .). In order to arrive at point ix at time (m + 1)t, theparticle must have been at one of the adjacent points (i ± 1)x at the preced-ing time mt and must have moved toward ix. From this, we see that the

2.13 Comments and References 205

i

m −3 −2 −1 0 1 2 30 0 0 0 1 0 01 0 0 0.5 0 0.5 0 02 0 0.25 0 0.5 0 0.25 03 0.125 0 0.375 0 0.375 0 0.125

Table 3 Random-walk probabilities

probabilities are related by the equation

ui(m + 1) = 1

2ui−1(m) + 1

2ui+1(m).

The u’s are completely determined once an initial probability distribution isgiven. For instance, if the particle is initially at point zero (u0(0) = 1, ui(0) = 0,for i �= 0), the ui(m) are formed by successive applications of the differenceequation, as shown in Table 3.

The equation may be transformed into a close relative of the heat equation.First, subtract ui(m) from both sides:

ui(m + 1) − ui(m) = 1

2

(ui+1(m) − 2ui(m) + ui−1(m)

).

Next divide by x2/2 on the right and by t · (x2/2t) on the left to obtain

ui(m + 1) − ui(m)

t

2t

(x)2= ui+1(m) − 2ui(m) + ui−1(m)

(x)2.

If both the time interval t and the step length x are small, we may thinkof ui(m) as being the value of a continuous function u(x, t) at x = ix, t =mt. In the limit, the difference quotient on the left approaches ∂u/∂t. Theright-hand side, being a difference of differences, approaches ∂2u/∂x2. Theheat equation thus results if, in the simultaneous limit as x and t tendto zero, the quantity 2t/(x)2 approaches a finite, nonzero limit. In thiscontext, the heat equation is called the Fokker–Planck equation. More detailsand references may be found in Feller, Introduction to Probability Theory andIts Applications.

We have used the term linear partial differential equation several times. Themost general such equation, of second order in two independent variables, canbe put in the form

A∂2u

∂x2+ B

∂2u

∂x∂t+ C

∂2u

∂t2+ D

∂u

∂x+ E

∂u

∂t+ Fu + G = 0,

where A,B, . . . ,G are known — perhaps functions of x and t but not of u orits derivatives. If G is identically zero, the equation is homogeneous. Of course,


the ordinary heat equation has this form if we take A = 1, E = −1/k and allother coefficients equal to zero.

Some astute students will have wondered why we should seek solutions inproduct form. The simplest answer is that in many cases it works. A moresubtle rationale is that of seeking solutions that are geometrically similar func-tions of x at different times. The idea of similarity — related to dimensionalanalysis — has been most fruitful in the mechanics of fluids.

Chapter ReviewSee the CD for Review Questions.

Miscellaneous ExercisesAlso see Review Questions on the CD.

In Exercises 1–16, find the steady-state solution, the associated eigenvalueproblem, and the complete solution for each problem.

1.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0, u(a, t) = T0, 0 < t,

u(x,0) = T1, 0 < x < a.

2.∂2u

∂x2− γ 2u = 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0, u(a, t) = T0, 0 < t,

u(x,0) = T1, 0 < x < a.

3.∂2u

∂x2+ r = 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0, u(a, t) = T0, 0 < t,

u(x,0) = T1, 0 < x < a (r is constant).

4.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0,∂u

∂x(a, t) = 0, 0 < t,

u(x,0) = T1x

a, 0 < x < a.

5.∂2u

∂x2− γ 2u = 1

k

∂u

∂t, 0 < x < a, 0 < t,


∂u

∂x(0, t) = 0,

∂u

∂x(a, t) = 0, 0 < t,

u(x,0) = T1x

a, 0 < x < a.

6.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = T0, 0 < t,

u(x,0) = 0, 0 < x < a.

7.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0, u(a, t) = T0,

u(x,0) = T0, 0 < x < a.

8.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

∂u

∂x(0, t) = T

a,

∂u

∂x(a, t) = T

a, 0 < t,

u(x,0) = T0, 0 < x < a.

9.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0,∂u

∂x(a, t) = 0, 0 < t,

u(x,0) = T1, 0 < x < a.

10.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

∂u

∂x(0, t) = 0,

∂u

∂x(a, t) = 0, 0 < t,

u(x,0) =

T0, 0 < x <a

2,

T1,a

2< x < a.

11.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < ∞, 0 < t,

u(0, t) = T0, 0 < t,

u(x,0) = T0

(1 − e−αx

), 0 < x.

12.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < ∞, 0 < t,


u(0, t) = T0, 0 < t,

u(x,0) ={0, 0 < x < a,

T0, a < x.

13.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < ∞, 0 < t,

∂u

∂x(0, t) = 0, 0 < t,

u(x,0) ={

T0, 0 < x < a,0, a < x.

14.∂2u

∂x2= 1

k

∂u

∂t, −∞ < x < ∞, 0 < t,

u(x,0) = exp(−α|x|), −∞ < x < ∞.

15.∂2u

∂x2= 1

k

∂u

∂t, −∞ < x < ∞, 0 < t,

u(x,0) ={0, −∞ < x < 0,

T0, 0 < x < a,0, a < x < ∞.

16.∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

∂u

∂x(0, t) = 0, u(a, t) = T0, 0 < t,

u(x,0) = T0 + S(a − x), 0 < x < a.

17. Give a physical interpretation for this problem and thus explain whyu(x, t) should increase steadily as t increases. (Assume that S is a pos-itive constant.)

∂2u

∂x2= 1

k

∂u

∂t, 0 < x < a, 0 < t,

∂u

∂x(0, t) = 0,

∂u

∂x(a, t) = S, 0 < t,

u(x,0) = 0, 0 < x < a.

18. Show that v(x, t) = (S/2a)(x2 + 2kt) satisfies the heat equation and theboundary conditions of the problem in Exercise 17. Also find w(x, t),defined by u(x, t) = v(x, t) + w(x, t).

19. Show that the four functions

u0 = 1, u1 = x, u2 = x2 + 2kt, u3 = x3 + 6kxt


are solutions of the heat equation. (These are sometimes called heatpolynomials.) Find a linear combination of them that satisfies theboundary conditions u(0, t) = 0, u(a, t) = t.

20. Suppose that u(x, t) is a positive function that satisfies

∂2u

∂x2= ∂u

∂t.

Show that the function

w(x, t) = −2

u

∂u

∂x

satisfies the nonlinear partial differential equation called Burgers’ equa-tion:

∂w

∂t+ w

∂w

∂x= ∂2w

∂x2.

21. Find a solution of the Burgers’ equation that satisfies the conditions

w(0, t) = 0, w(1, t) = 0, 0 < t,

w(x,0) = 1, 0 < x < 1.

22. Taking the function u(x, t) given here as a solution of the heat equation(with k = 1), find a solution w of Burgers’ equation. Verify that w satis-fies Burgers’ equation.

u(x, t) = 1√4π t

exp

(−x2

4t

).

23. Consider a solid metal bar surrounded by a finite quantity of water con-fined in a water jacket. If the bar and the water are at different tempera-tures, they will exchange heat. Let u1 and u2 be the temperatures in thebar and in the water, respectively. Heat balances for the water and the bargive these two equations:

c1du1

dt= h(u2 − u1),

c2du2

dt= h(u1 − u2).

Here, c1 and c2 are the heat capacities of the bar and the water, respec-tively, and h is the product of the convection coefficient with the areaof the bar–water interface. Find temperatures u1 and u2 assuming initialconditions u1(0) = T0, u2(0) = 0.


24. Solve the eigenvalue problem by setting φ(ρ) = ψ(ρ)/ρ:

1

ρ2

(ρ2φ′)′ + λ2φ = 0, 0 < ρ < a,

φ(0) bounded, φ(a) = 0.

Is this a regular Sturm–Liouville problem? Are the eigenfunctions or-thogonal?

25. Solve this problem for heat conduction in a sphere. (Hint: Let u(ρ, t) =v(ρ, t)/ρ.)

1

ρ2

∂

∂ρ

(ρ2 ∂u

∂ρ

)= 1

k

∂u

∂t, 0 < ρ < a, 0 < t,

u(0, t) bounded, u(a, t) = 0, 0 < t,

u(ρ,0) = T0, 0 < ρ < a.

26. State and solve the eigenvalue problem associated with

e−x ∂

∂x

(ex ∂u

∂x

)= 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = 0,∂u

∂x(a, t) = 0.

27. Find the steady-state solution of the problem

∂2u

∂x2+ γ 2

(T(x) − u

) = 1

k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = T0,∂u

∂x(a, t) = 0, 0 < t,

where T(x) = T0 + Sx.

28. Determine whether or not λ = 0 is an eigenvalue of the problem

φ′′ + λ2xφ = 0, 0 < x < a,

φ′(0) = 0, φ(a) = 0.

29. Same question as Exercise 28, but with boundary conditions

φ′(0) = 0, φ′(a) = 0.


30. Prove the following identity:

1√4πkt

∫ a

bexp

[− (ξ − x)2

4kt

]dξ = 1

2

[erf

(b − x√

4kt

)− erf

(a − x√

4kt

)].

31. In Exercise 6 of Section 10, it was shown that the function

w(x, t;ω) = e−px sin(ωt − px),

where p = √ω/2k, satisfies the heat equation and also the boundary con-

dition

w(0, t;ω) = sin(ωt).

Show how to choose the coefficient B(ω) so that the function

u(x, t) =∫ ∞

0B(ω)e−px sin(ωt − px)dω

satisfies the boundary condition

u(0, t) = f (t), 0 < t

for a suitable function t.

32. Use the idea of Exercise 31 to find a solution of

∂2u

∂x2= 1

k

∂u

∂t, 0 < x, 0 < t,

u(0, t) = h(t), 0 < t,

where

h(t) ={

1, 0 < t < T,0, T < t.

33. S.E. Serrano and T.E. Unny develop probabilistic mathematical modelsfor groundwater flow under uncertain conditions [Predicting groundwa-ter flow in a phreatic aquifer, Journal of Hydrology, 95 (1987): 241–268],and compare the results to measurements. One of the models uses thisnonlinear Boussinesq equation,

S∂y

∂t− ∂

∂x

(Kh

∂y

∂x

)= I + φ, 0 < x < L, 0 < t,

together with the conditions

y(0, t) = y1(t), y(L, t) = y2(t), 0 < t,

y(x,0) = y0(x), 0 < x < L.


In these equations, y(x, t) is the water table elevation above sea level,h(x, t) is water table elevation above bedrock, K is hydraulic conductiv-ity (in meters per day, m/da), S is the aquifer specific yield, I is a functionrepresenting input by percolation from the aquifer, and φ(x, t) is a ran-dom function that accounts for uncertainty in input.

The partial differential equation is nonlinear because h and y representthe same thing relative to two different references. We assume that thebedrock elevation has constant slope a, so y = h + ax. Then the equationcan be written in terms of h alone as

S∂h

∂t− ∂

∂x

(Kh

∂h

∂x

)− a

∂

∂x(Kh) = I + φ.

Next, this equation is linearized. Assume that K is constant and that hcan be broken down as h = h + h′, where h is a constant mean value of h,h′ is a fluctuation much smaller than h. (In this case, h is about 150 mand h′ is less than 1 m.) Then the product Kh is approximately equal toKh = T (transmissivity) and, as a coefficient in the second term, can betreated as a constant. The equation is now linear in h′ (we drop the primefor convenience):

S∂h

∂t− T

∂2h

∂x2− aK

∂h

∂x= I + φ, 0 < x < L, 0 < t.

a. Treating I as a constant, find a steady-state solution v(x) for the sta-tistical mean value of h, which is obtained by replacing φ(x, t) with 0.The boundary and initial conditions are

h(0, t) = h1, h(L, t) = h2, 0 < t,

h(x,0) = h0(x), 0 < x < L.

b. State the problem (partial differential equation, boundary conditions,and initial condition) to be satisfied by the mean transient, w(x, t) =h(x, t) − v(x). (Again, the statistical mean corresponds to φ ≡ 0.)

c. Solve the problem in b.

d. Values for the parameters are: a = 0.0292 m/m, K = 17.28 m/da, T =218.4 m2/da, S = 0.15, L = 116.25 m. Find the eigenvalues.

34. A flat enzyme electrode can be visualized by imagining it seen from theside. The electrode itself lies to the left of x = 0 (its thickness is unimpor-tant); a gel-containing enzyme lies in a layer between x = 0 and x = L;and the test solution lies to the right of x = L. When the substance to bedetected is introduced into the solution, it diffuses into the gel and re-acts with the enzyme, yielding a product. The electrode responds to theproduct with a measurable electric potential.


P.W. Carr [Fourier analysis of the transient response of potentiomet-ric enzyme electrodes, Analytical Chemistry, 49 (1977): 799–802] stud-ied the transient response of such an electrode via two partial differentialequations that describe the concentrations, S and P, of the substance be-ing detected and the enzyme-reaction product as they diffuse and reactin the gel:

∂S

∂t= D

∂2S

∂x2− VS

K + S, 0 < x < L, 0 < t, (1∗)

∂P

∂t= D

∂2P

∂x2+ VS

K + S, 0 < x < L, 0 < t. (2∗)

In these equations, V is the specific enzyme activity (mol/ml s), K is aconstant related to reaction rate, and D is the diffusion constant (cm2/s),assumed to be the same for both substance and product.

Reasonable boundary conditions are

∂S

∂x(0, t) = 0,

∂P

∂x(0, t) = 0, 0 < t, (3∗)

representing no reaction or penetration at the electrode surface, and

S(L, t) = S0, P(L, t) = 0, 0 < t, (4∗)

where the gel meets the test solution. Initially, we assume

S(x,0) = 0, P(x,0) = 0, 0 < x < L. (5∗)

Equation (1∗) is nonlinear because the unknown function S appears inthe denominator of the last term. However, if S is much smaller than K ,we may replace K + S by K , and Eq. (1∗) becomes

∂S

∂t= D

∂2S

∂x2− V

KS, 0 < x < L, 0 < t. (6∗)

a. State and solve the steady-state problem for this equation, subject tothe boundary conditions on S in Eqs. (3∗) and (4∗).

b. Find the transient solution and then the complete solution S(x, t).

35. Refer to Exercise 34. Equation (2∗), though linear, is not easy to solve.However, if Eqs. (1∗) and (2∗) are added together, the nonlinear termscancel, leaving this homogeneous linear equation for the sum of the con-centrations:

∂(S + P)

∂t= D

∂2(S + P)

∂x2, 0 < x < L, 0 < t.


Defining u = S + P, find the boundary and initial conditions for u, andsolve completely. Then find P(x, t) as u(x, t) − S(x, t).

36. Refer to Exercises 34 and 35. In order to determine the response timeof the enzyme electrode, one wants to know the function P(0, t). Ap-proximate this, using in your solution only steady-state terms and thefirst term of each infinite series. Sketch. Find the “time constants,” themultipliers of t in the exponential functions.

37. Consider a steel plate that is much larger in length and width (x- and z-directions) than in thickness (y-direction), and suppose the plate is freeto expand or contract under the effects of heating. Assume that the tem-perature T in the plate is a function of y and t only. Timoshenko andGoodier (Theory of Elasticity, pp. 399–403) derive the following expres-sion for the stresses due to thermal effects:

σx = σz = − αTE

1 − ν+ 1

2c(1 − ν)

∫ +c

−cαTE dy

+ 3y

2c3(1 − ν)

∫ +c

−cαTEy dy.

The parameters, and their values for steel are as follows: α is the co-efficient of expansion, 6.5 × 10−6 per degree F; E is Young’s modulus,28 × 106 lb/in.2; ν is Poisson’s ratio, 0.7; and 2c is the thickness of theplate. Note that the origin is located so that the plate lies between y = cand y = −c.

a. Show that if T(y) = T0 + Sy, where T0 and S are constants, then thethermal stress is 0. (This is a typical steady-state temperature distri-bution.)

b. Suppose that the plate is initially at temperature 500◦F throughoutand that the temperature on the face y = c is suddenly changed to200◦ while the temperature at y = −c remains at 500◦. Find T(y, t).

c. Assume the initial and boundary conditions given in b. Use your un-derstanding of the function T(y, t) to explain why the thermal stressnear the face y = c is large just after time 0.

The Wave EquationC H A P T E R

3

3.1 The Vibrating String

A simple and historically important example of a problem that includes thewave equation is provided by the study of the vibration of a string, like a violinor guitar string. We set up a coordinate system as shown in Fig. 1. The un-known is the transverse displacement, u(x, t), measured up from the x-axis.The situation is similar to that of the hanging cable discussed in Chapter 0,but here the string is taut, and of course motion is allowed. In order to find theequation of motion of the string, we consider a short piece whose ends are atx and x + x and apply Newton’s second law of motion to it.

First, we must analyze the nature of the forces on the string. We assumethat the only external force is the attraction of gravity, acting perpendicularto the x-direction. Internal forces are exerted on the segment by the rest of thestring. We will assume that the string is perfectly flexible and offers no resistanceto bending. Then the only force that can be transmitted by the string is a pullor tension, which acts in a direction tangential to the centerline of the string.Its magnitude we denote by T(x, t).

The forces on the small segment of string are shown in Fig. 2. We shall fur-ther assume that each point on the string moves only in the vertical direction.Thus, the horizontal component of acceleration is zero. Application of New-ton’s second law for the horizontal direction to the segment leads to the equa-tion

−T(x, t) cos(φ(x, t)

) + T(x + x, t) cos(φ(x + x, t)

) = 0,

215

216 Chapter 3 The Wave Equation

Figure 1 String fixed at the ends.

Figure 2 Section of string showing forces exerted on it. The angles areα = φ(x, t) and β = φ(x + x, t).

or

T(x, t) cos(φ(x, t)

) = T(x + x, t) cos(φ(x + x, t)

). (1)

This says that the horizontal component of tension in the string is the same atevery point:

T(x, t) cos(φ(x, t)

) = T(x + x, t) cos(φ(x + x, t)

) = T,

independent of x. If the string is taut, T can vary only slightly with t, so we willassume that T is constant.

In the absence of external forces other than gravity, Newton’s second law forthe vertical direction yields

−T(x, t) sin(φ(x, t)

) + T(x + x, t) sin(φ(x + x, t)

) − mg = m∂2u

∂t2(x, t).

(2)

(Because u(x, t) measures displacement in the vertical direction, its secondpartial derivative with respect to t is the vertical acceleration.) The mass of theshort piece of string we are examining is proportional to its length, m = ρ x,where ρ is the linear density, measured in units of mass per unit length.

Now we use Eq. (1) to solve for the tensions at the ends of the segment ofstring as

T(x, t) = T

cos(φ(x, t)), T(x + x, t) = T

cos(φ(x + x, t)).

Chapter 3 The Wave Equation 217

When these expressions are substituted into Eq. (2), we have

−T tan(φ(x, t)

) + T tan(φ(x + x, t)

) − ρ x g = ρ x∂2u

∂t2. (3)

Recall from elementary calculus that tan(φ(x, t)) is the slope of the string at(x, t) and hence may be expressed in terms of the (partial) derivative withrespect to x:

tan(φ(x, t)

) = ∂u

∂x(x, t), tan

(φ(x + x, t)

) = ∂u

∂x(x + x, t).

Substituting these into Eq. (3), we have

T

(∂u

∂x(x + x, t) − ∂u

∂x(x, t)

)= ρx

(∂2u

∂t2+ g

).

On dividing through by x, we see a difference quotient on the left:

T

x

(∂u

∂x(x + x, t) − ∂u

∂x(x, t)

)= ρ

(∂2u

∂t2+ g

).

In the limit as x → 0, the difference quotient becomes a partial derivativewith respect to x, leaving Newton’s second law in the form

T∂2u

∂x2= ρ

∂2u

∂t2+ ρg, (4)

or

∂2u

∂x2= 1

c2

∂2u

∂t2+ 1

c2g, (5)

where c2 = T/ρ. If c2 is very large (usually on the order of 105 m2/s2), weneglect the last term, giving the equation of the vibrating string

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t. (6)

This equation is called the wave equation in one dimension. Two- and three-dimensional versions will be treated in Chapter 5.

In describing the motion of an object, one must specify not only the equa-tion of motion, but also both the initial position and velocity of the object.The initial conditions for the string, then, must state the initial displacementof every particle — that is, u(x,0) — and the initial velocity of every particle,∂u/∂t(x,0).

For the vibrating string as we have described it, the boundary conditions arezero displacement at the ends, so the boundary value–initial value problem for


the string is

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t, (7)

u(0, t) = 0, u(a, t) = 0, 0 < t, (8)

u(x,0) = f (x), 0 < x < a, (9)

∂u

∂t(x,0) = g(x), 0 < x < a, (10)

under the assumptions noted plus the assumption that gravity is negligible.

E X E R C I S E S

1. Find the dimensions of each of the following quantities, using the facts thatforce is equivalent to mL/t2, and that the dimension of tension is F (force):u, ∂2u/∂x2, ∂2u/∂t2, c, g/c2. Check the dimension of each term in Eq. (5).

2. Suppose a distributed vertical force F(x, t) (positive upwards) acts on thestring. Derive the equation of motion:

∂2u

∂x2= 1

c2

∂2u

∂t2− 1

TF(x, t).

The dimension of a distributed force is F/L. (If the weight of the string isconsidered as a distributed force and is the only one, then we would haveF(x, t) = −ρg. Check dimensions and signs.)

3. Find a solution v(x) of Eq. (5) with boundary conditions Eq. (8) that isindependent of time. (This corresponds to a “steady-state solution,” butthe term steady-state is no longer appropriate. Equilibrium solution is moreaccurate.)

4. Suppose that the string is located in a medium that resists its movement,such as air. The resistance is expressed as a force opposite in direction andproportional in magnitude to velocity. Thus it affects only Eq. (2). Proceedto derive the equation that replaces Eq. (7) for this case.

3.2 Solution of the Vibrating String Problem

The initial value–boundary value problem that describes the displacement ofthe vibrating string,

3.2 Solution of the Vibrating String Problem 219

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t, (1)

u(0, t) = 0, u(a, t) = 0, 0 < t, (2)

u(x,0) = f (x), 0 < x < a, (3)

∂u

∂t(x,0) = g(x), 0 < x < a, (4)

contains a linear, homogeneous partial differential equation and linear, homo-geneous boundary conditions. Thus we may apply the method of separation ofvariables with hope of success. If we assume that1 u(x, t) = φ(x)T(t), Eq. (1)becomes

φ′′(x)T(t) = 1

c2φ(x)T ′′(t).

Dividing through by φT, we obtain

φ′′(x)

φ(x)= T ′′(t)

c2T(t), 0 < x < a, 0 < t.

For the equality to hold, both members of this equation must be constant.We write the constant as −λ2 and separate the preceding equation into twoordinary differential equations linked by the common parameter λ2:

T ′′ + λ2c2T = 0, 0 < t, (5)

φ′′ + λ2φ = 0, 0 < x < a. (6)

The boundary conditions become

φ(0)T(t) = 0, φ(a)T(t) = 0, 0 < t

and, since T(t) ≡ 0 gives a trivial solution for u(x, t), we must have

φ(0) = 0, φ(a) = 0. (7)

The eigenvalue problem Eqs. (6) and (7) is exactly the same as the one wehave seen and solved before. (See Chapter 2, Section 3.) We know that theeigenvalues and eigenfunctions are

λ2n =

(nπ

a

)2

, φn(x) = sin(λnx), n = 1,2,3, . . . .

Equation (5) is also of a familiar type, and its solution is known to be

Tn(t) = an cos(λnct) + bn sin(λnct),

1T no longer symbolizes tension.


where an and bn are arbitrary. (In other words, there are two independent solu-tions.) Note, however, that there is a substantial difference between the T thatarises here and the T that we found in the heat conduction problem. The mostimportant difference is the behavior as t tends to infinity. In the heat conduc-tion problem, T(t) tends to 0, whereas here T(t) has no limit but oscillatesperiodically in agreement with our intuition.

For each n = 1,2,3, . . . , we now have product solutions

un(x, t) = sin(λnx)[an cos(λnct) + bn sin(λnct)

]. (8)

Such solutions are called standing waves. For a particular an and bn, un(x, t)maintains the same shape with a variable, periodic amplitude. For any choiceof an and bn, un(x, t) is a solution of the homogeneous partial differentialequation (1) and also satisfies the boundary conditions Eq. (2). Some standingwaves are shown animated on the CD.

By the Principle of Superposition, linear combinations of the un(x, t) alsosatisfy both Eqs. (1) and (2). In making our linear combinations, we need nonew constants because the an and bn are arbitrary. We have, then,

u(x, t) =∞∑

n=1

sin(λnx)[an cos(λnct) + bn sin(λnct)

]. (9)

The initial conditions, which remain to be satisfied, have the form

u(x,0) =∞∑

n=1

an sin

(nπx

a

)= f (x), 0 < x < a,

∂u

∂t(x,0) =

∞∑n=1

bnnπ

ac sin

(nπx

a

)= g(x), 0 < x < a.

(Here we have assumed that

∂u

∂t(x, t) =

∞∑n=1

sin(λnx)[−anλnc sin(λnct) + bnλnc cos(λnct)

].

In other words, we assume that the series for u may be differentiated term byterm.) Both initial conditions take the form of Fourier series problems: A givenfunction is to be expanded in a series of sines. In each case, then, the constantmultiplying sin(nπx/a) must be the Fourier sine coefficient for the given func-tion. Thus we determine that

an = 2

a

∫ a

0f (x) sin

(nπx

a

)dx, (10)


and

bnnπ

ac = 2

a

∫ a

0g(x) sin

(nπx

a

)dx

or

bn = 2

nπc

∫ a

0g(x) sin

(nπx

a

)dx. (11)

If the functions f (x) and g(x) are sectionally smooth on the interval 0 < x <

a, then we know that the initial conditions really are satisfied, except possiblyat points of discontinuity of f or g. By the nature of the problem, however, onewould expect that f , at least, would be continuous and would satisfy f (0) =f (a) = 0. Thus we expect the series for f to converge uniformly.

Example.If the string is lifted in the middle and then released, appropriate initial condi-tions are

u(x,0) = f (x) =

h · 2x

a, 0 < x <

a

2,

h

(2 − 2x

a

),

a

2< x < a,

∂u

∂t(x,0) = g(x) ≡ 0, 0 < x < a.

Then bn = 0 for n = 1,2,3, . . . , and

an = 2

a

[∫ a/2

0h · 2x

asin

(nπx

a

)dx +

∫ a

a/2h

(2 − 2x

a

)sin

(nπx

a

)dx

]

= 8h

π2

sin(nπ/2)

n2.

Therefore the complete solution is

u(x, t) = 8h

π2

∞∑n=1

sin(nπ/2)

n2sin

(nπx

a

)cos

(nπct

a

). (12)

The CD shows an animated version of this solution. �

Although the solution in the example can be considered valid, it is difficultto see, in the present form, what shape the string will take at various times.However, because of the simplicity of the sines and cosines, it is possible torewrite the solution in such a way that u(x, t) may be determined withoutsumming a series.


By applying the trigonometric identity

sin(A) cos(B) = 1

2

[sin(A − B) + sin(A + B)

]we can express u(x, t) as

u(x, t) = 1

2

[8h

π2

∞∑n=1

sin(nπ/2)

n2sin

(nπ(x − ct)

a

)

+ 8h

π2

∞∑n=1

sin(nπ/2)

n2sin

(nπ(x + ct)

a

)].

We know that the series

8h

π2

∞∑n=1

sin(nπ/2)

n2sin

(nπx

a

)

actually converges to the odd periodic extension, with period 2a, of the func-tion f (x). Let us designate this extension by fo(x) and note that it is defined forall values of its argument. Using this observation, we can express u(x, t) moresimply as

u(x, t) = 1

2

[fo(x − ct) + fo(x + ct)

]. (13)

In this form, the solution u(x, t) can easily be sketched for various valuesof t. The graph of fo(x + ct) has the same shape as that of fo(x) but is shifted ctunits to the left. Similarly, the graph of fo(x − ct) is the graph of fo(x) shiftedct units to the right. When the graphs of fo(x + ct) and fo(x − ct) are drawn onthe same axes, they may be averaged graphically to get the graph of u(x, t).

In Fig. 3 are graphs of fo(x + ct), fo(x − ct), and

u(x, t) = 1

2

[fo(x + ct) + fo(x − ct)

]for the particular example discussed here and for various values of t. The dis-placement u(x, t) is periodic in time, with period 2a/c. During the secondhalf-period (not shown), the string returns to its initial position through thepositions shown. The horizontal portions of the string have a nonzero veloc-ity. Equation (12) can also be used to find u(x, t) for any given x and t. Forinstance, if we take x = 0.2a and t = 0.9a/c, we find that

u

(0.2a,0.9

a

c

)= 1

2

[fo(−0.7a) + fo(1.1a)

]= 1

2

[(−0.6h) + (−0.2h)

]= −0.4h.


Figure 3 On the left are the graphs of fo(x + ct) (solid) and fo(x − ct) (dashed)for the given value of ct. On the right is the graph of u(x, t) for 0 < x < a, madeby averaging the graphs on the left.

The function values can be read directly from a graph of f (x). The manipula-tions with the series solution in the example can be done for any f (x). There-fore the formula of Eq. (13) is a solution of Eqs. (1)–(4) for any f (x), providedthat g(x) ≡ 0. We will generalize in later sections.

Frequencies of VibrationThe product solutions in Eq. (8) provide important information about pos-sible frequencies of vibration. The multipliers λnc in the sines and cosines of


t are frequencies, in radians per unit time; λnc/2π are frequencies in cyclesper unit time (or Hertz, if the time unit is seconds). For the vibrating stringproblem, the possible frequencies of vibration are

(nπ/a)c

2π= n

πc

2a.

The fact that these form an arithmetic sequence guarantees a common periodfor all the un(x, t) in Eq. (8), and thus u(x, t) in Eq. (9) is a function that isperiodic in time.

E X E R C I S E S

1. Verify that the product solution

un(x, t) = sin(λnx)[an cos(λnct) + bn sin(λnct)

]satisfies the wave equation (1) and the boundary conditions, Eq. (2).

2. Sketch u1(x, t) and u2(x, t) as functions of x for several values of t. Assumea1 and a2 = 1, b1 and b2 = 0. (The solutions un(x, t) are called standingwaves.)

In Exercises 3–5, solve the vibrating string problem, Eqs. (1)–(4), with the ini-tial conditions given.

3. f (x) = 0, g(x) = 1, 0 < x < a.

4. f (x) = sin

(πx

a

), g(x) = 0, 0 < x < a.

5. f (x) ={

U0, 0 < x < a/2,0, a/2 < x < a,

g(x) = 0, 0 < x < a.

(This initial condition is difficult to justify for a vibrating string, but itmay be reasonable where the unknown function is pressure in a pipe witha membrane at the midpoint. See Miscellaneous Exercise 18 of this chapterfor some derivations.)

6. If

∞∑n=1

an sin

(nπx

a

)= fo(x),

∞∑n=1

bn cos

(nπx

a

)= Ge(x),

show that u(x, t) as given in Eq. (9) may be written


u(x, t) = 1

2

(fo(x − ct) + fo(x + ct)

) + 1

2

(Ge(x + ct) − Ge(x − ct)

).

Here, fo(x) and Ge(x) are periodic with period 2a.

7. The pressure of the air in an organ pipe satisfies the equation

∂2p

∂x2= 1

c2

∂2p

∂t2, 0 < x < a, 0 < t,

with the boundary conditions (p0 is atmospheric pressure)

a. p(0, t) = p0, p(a, t) = p0 if the pipe is open, or

b. p(0, t) = p0, ∂p∂x (a, t) = 0 if the pipe is closed at x = a.

Find the eigenvalues and eigenfunctions associated with the wave equationfor each of these sets of boundary conditions.

8. Find the lowest frequency of vibration of the air in the organ pipes referredto in Exercise 7a and b.

9. If a string vibrates in a medium that resists the motion, the problem forthe displacement of the string is

∂2u

∂x2= 1

c2

∂2u

∂t2+ k

∂u

∂t, 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = 0, 0 < t

plus initial conditions. Find eigenfunctions, eigenvalues, and product so-lutions for this problem. (Assume that k is small and positive.)

10. For the problem in Exercise 9, find frequencies of vibration and show thatthey do not form an arithmetic sequence. If we form a series solution, willit be periodic? What happens to u(x, t) as t → ∞?

11. The displacements u(x, t) of a uniform thin beam satisfy

∂4u

∂x4= − 1

c2

∂2u

∂t2, 0 < x < a, 0 < t.

If the beam is simply supported at the ends, the boundary conditions are

u(0, t) = 0,∂2u

∂x2(0, t) = 0, u(a, t) = 0,

∂2u

∂x2(a, t) = 0.

Find product solutions to this problem. What are the frequencies of vibra-tion?

12. Write out formulas for the first four frequencies of vibration for a thinbeam (Exercise 11) and for a string (text). Then find their values, assum-ing that parameters c and a have values that make the lowest frequency of


Figure 4 Shapes of car antenna.

each equal to 256 cycles per second. The difference in the set of frequen-cies accounts for some of the difference between the sound of a stringedinstrument and that of a xylophone or glockenspiel.

13. My car’s antenna vibrates in the wind under various conditions in one ofthe two shapes shown in Fig. 4. If the antenna is modeled as a uniform thinbeam with centerline displacement u(x, t), then u satisfies the equation

∂4u

∂x4= − 1

c2

∂2u

∂t2+ f (x, t), 0 < x < a, 0 < t,

where f is a “forcing function” that represents the effect of wind or otherdistributed forces. Because the base of the antenna is built into the car, theboundary conditions at the base are zero displacement and slope:

u(0, t) = 0,∂u

∂x(0, t) = 0, 0 < t.

The top of the antenna is free to move. There, the internal moment andshear are both zero, leading to the conditions

∂2u

∂x2(a, t) = 0,

∂3u

∂x3(a, t) = 0, 0 < t.

(These four boundary conditions are standard for a cantilevered beam.)It can be shown that the solution of the foregoing problem, together with

initial conditions on u and ut , can be represented as a series of productsof the form (

an cos(λ2

nct) + bn sin

(λ2

nct) + Fn(t)

)φn(x),

where Fn(t) comes from the forcing function and φn(x) and λn are relatedthrough the eigenvalue problem

φ′′′′ − λ4φ = 0, 0 < x < a,

φ(0) = 0, φ′(0) = 0, φ′′(a) = 0, φ′′′(a) = 0.

This arises in the obvious way from the boundary conditions and the ho-mogeneous partial differential equation.

3.3 d’Alembert’s Solution 227

Solve the eigenvalue problem, sketch the first two eigenfunctions, andcompare them to the figure.

In Exercises 14–16, find a solution by separation of variables.

14.∂2u

∂x2= 1

c2

(∂2u

∂t2+ 2k

∂u

∂t

), 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = 0, 0 < t,

u(x,0) = f (x),∂u

∂t(x,0) = 0, 0 < x < a,

where f (x) is as in Eq. (11). (Assume that k is small and positive.)

15.∂2u

∂x2= 1

c2

∂2u

∂t2+ γ 2u, 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = 0, 0 < t,

u(x,0) = h,∂u

∂t(x,0) = 0, 0 < x < a,

where h and γ 2 are constants.

16.∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t,

u(0, t) = 0,∂u

∂x(a, t) = 0, 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 1, 0 < x < a.

17. Does the series in Eq. (12) converge uniformly?

18. In the text, we assumed that the ratio φ′′/φ had to be a negative con-stant. Show that, if φ′′/φ = p2 > 0 (or equivalently, if φ′′ − p2φ = 0), thenthe only function that also satisfies the boundary conditions, Eq. (7), isφ(x) ≡ 0.

3.3 d’Alembert’s SolutionIn Section 2 we saw that, in some cases, we could express the solution of thewave equation directly in terms of the initial data. From this evidence we mightsuspect that there is something special about x + ct and x − ct. To test this idea,we change variables and see what the wave equation looks like. Let w = x + ct,z = x− ct, and u(x, t) = v(w, z). Then a calculation using the chain rule showsthat the wave equation becomes (see Exercise 11)

∂2v

∂z ∂w= 0.


It is actually possible to find the general solution of this last equation. Put inanother form it says

∂

∂z

(∂v

∂w

)= 0,

which means that ∂v/∂w is independent of z, or

∂v

∂w= θ(w).

Integrating this equation, we find that

v =∫

θ(w)dw + φ(z).

Here, φ(z) plays the role of an integration “constant.” Since the integral ofθ(w) is also a function of w, we may write the general solution of the partialdifferential equation foregoing as

v(w, z) = ψ(w) + φ(z),

where ψ and φ are arbitrary functions with continuous derivatives. Trans-forming back to our original variables, we obtain

u(x, t) = ψ(x + ct) + φ(x − ct) (1)

as a form for the general solution of the one-dimensional wave equation. Thisis known as d’Alembert’s solution or the traveling wave solution. It representsthe solution as the superposition of two waves, one moving to the left and theother to the right, with propagation speed c.

Now let us look at the vibrating string problem:

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t, (2)

u(0, t) = 0, u(a, t) = 0, 0 < t, (3)

u(x,0) = f (x), 0 < x < a, (4)

∂u

∂t(x,0) = g(x), 0 < x < a. (5)

We already know a form for u. The problem is to choose ψ and φ in such a waythat the initial and boundary conditions are satisfied. We assume then that

u(x, t) = ψ(x + ct) + φ(x − ct).

The initial conditions are

ψ(x) + φ(x) = f (x), 0 < x < a,

cψ ′(x) − cφ′(x) = g(x), 0 < x < a.(6)


If we divide through the second equation by c and integrate, it becomes

ψ(x) − φ(x) = G(x) + A, 0 < x < a, (7)

where G(x) stands for

G(x) = 1

c

∫ x

0g(y)dy (8)

and A is an arbitrary constant. Equations (6) and (7) can now be solved simul-taneously to determine

ψ(x) = 1

2

(f (x) + G(x) + A

), 0 < x < a,

φ(x) = 1

2

(f (x) − G(x) − A

), 0 < x < a.

These equations give ψ and φ only for values of the argument between 0and a. But x ± ct may take on any value whatsoever, so we must extend thesefunctions to define them for arbitrary values of their arguments. Let us desig-nate them as

ψ(x) = 1

2

(f (x) + G(x) + A

),

φ(x) = 1

2

(f (x) − G(x) − A

),

where f and G are some extensions of f and G. (That is f (x) = f (x) and G(x) =G(x) for 0 < x < a.) However we choose these extensions, the wave equationand the initial conditions are satisfied. Thus, they must be determined by theboundary conditions,

u(0, t) = ψ(ct) + φ(−ct) = 0, t > 0, (9)

u(a, t) = ψ(a + ct) + φ(a − ct) = 0, t > 0. (10)

The first of these equations says that

f (ct) + G(ct) + A + f (−ct) − G(−ct) − A = 0,

or

f (ct) + f (−ct) + G(ct) − G(−ct) = 0.

As these equations must be true for arbitrary functions f and G (because thetwo functions are not interdependent), we must have individually

f (ct) = −f (−ct), G(ct) = G(−ct). (11)

That is, f is odd and G is even.


At the second endpoint, a similar calculation shows that

f (a + ct) + f (a − ct) + G(a + ct) − G(a − ct) = 0.

Once again, the independence of f and G implies that

f (a + ct) = −f (a − ct), G(a + ct) = G(a − ct). (12)

The oddness of f and evenness of G can be used to transform the right-handsides. Then

f (a + ct) = f (−a + ct), G(a + ct) = G(−a + ct).

These equations say that f and G are both periodic with period 2a, becausechanging their arguments by 2a does not change the functional value. Thus

we want f to be the odd periodic extension of f and G to be the even periodicextension of G. In the notation we used in Chapter 1, the explicit expressionsfor φ and ψ are

ψ(x + ct) = 1

2

(fo(x + ct) + Ge(x + ct) + A

),

φ(x − ct) = 1

2

(fo(x − ct) − Ge(x − ct) − A

).

Finally, we arrive at an expression for the solution u(x, t):

u(x, t) = 1

2

[fo(x + ct) + fo(x − ct)

] + 1

2

[Ge(x + ct) − Ge(x − ct)

]. (13)

The CD shows an animated version of Fig. 3 (Section 3.2) using this form ofthe solution.

This form of the solution of Eqs. (2)–(5) allows us to see directly how theinitial data influence the solution at later times. From a practical point of view,it also permits us to calculate u(x, t) at any x and t and even to sketch u as afunction of one variable for a fixed value of the other. The following procedureis helpful in sketching u(x, t) as a function of x for a fixed t = t∗, when theinitial condition (5) has g(x) ≡ 0. It is easily adapted to other cases.

1. Sketch the odd periodic extension of f ; call this fo(x).

2. Sketch fo(x + ct∗) against x; this is just the graph of fo shifted ct∗ units tothe left.

3. Sketch fo(x − ct∗) against x on the same axes. This graph is the same asthat of fo but shifted ct∗ units to the right.

4. Average graphically the graphs made in the two preceding steps. Checkthat the boundary conditions are satisfied.


Similarly, if f (x) ≡ 0, sketch G(x) and its even periodic extension Ge(x).Then sketch the graphs of Ge(x + ct∗) (same shape as Ge(x) but shifted ct∗units to the left) and −Ge(x − ct∗) (graph of Ge(x) shifted ct∗ units to theright and reflected in the horizontal axis). These two are then averaged graph-ically to obtain the graph of u(x, t∗). Check that the boundary conditions aresatisfied.

E X E R C I S E S

1. Let u(x, t) be a solution of Eqs. (2)–(5), with g(x) ≡ 0 and f (x) a functionwhose graph is an isosceles triangle of width a and height h. Find u(x, t)for x = 0.25a and 0.5a and for t = 0, 0.2a/c, 0.4a/c, 0.8a/c, 1.4a/c.

2. Sketch u(x, t) of Exercise 1 as a function of x for the times given. Compareyour results with Fig. 3.

3. Let u(x, t) be a solution of Eqs. (2)–(5), with f (x) ≡ 0 and g(x) = αc, 0 <

x < a. Find u(x, t) at: x = 0, t = 0.5a/c; x = 0.2a, t = 0.6a/c; x = 0.5a,t = 1.2a/c.

4. Sketch u(x, t) of Exercise 3 as a function of x for times t = 0, 0.25a/c,0.5a/c, a/c.

5. Find the function G(x) corresponding to (see Eq. (8))

g(x) ={0, 0 < x < 0.4a,

5c, 0.4a < x < 0.6a,0, 0.6a < x < a.

6. Justify this alternate description of the function G(x), that is specified inEq. (8): G is the solution of the initial value problem

dG

dx= 1

cg(x), 0 < x,

G(0) = 0.

7. Using Eq. (8) or Exercise 6, sketch the function G(x) of Exercise 5.

8. Let u(x, t) be the solution of the vibrating string problem, Eqs. (2)–(5),with f (x) ≡ 0 and g(x) as in Exercise 5. Sketch u(x, t) as a function ofx for times ct = 0, 0.2a, 0.4a, 0.5a, a, 1.2a. Hint: Sketch Ge(x + ct) and−Ge(x − ct); then average them graphically.


9. Sketch the solution of the vibrating string problem, Eqs. (2)–(5), at timesct = 0, 0.1a, 0.3a, 0.4a, 0.5a, 0.6a, if g(x) = 0 and

f (x) =

0, 0 < x < 0.4a,10h(x − 0.4a), 0.4a < x < 0.5a,10h(0.6a − x), 0.5a < x < 0.6a,0, 0.6a < x < a.

10. Verify directly that u(x, t) as given by Eq. (1) is a solution of the waveequation (2) if φ and ψ have at least two derivatives.

11. Use the change of variables at the beginning of this section to transformthe wave equation. You need to use the chain rule extensively — for in-stance,

∂u

∂t= ∂v

∂w

∂w

∂t+ ∂v

∂z

∂z

∂t= ∂v

∂w· c + ∂v

∂z· (−c).

In this way, find expressions for the second derivatives of u, and then sub-stitute into the wave equation,

∂2u

∂x2= 1

c2

∂2u

∂t2.

12. The equation for the forced vibrations of a string is

∂2u

∂x2− 1

c2

∂2u

∂t2= − 1

TF(x, t) (∗)

(see Section 1, Exercise 2). Changing variables to

w = x + ct, z = x − ct, u(x, t) = v(w, z), f (w, z) = F(x, t),

this equation becomes

∂2v

∂w∂z= − 1

4Tf (w, z).

Show that the general solution of this equation is

v(w, z) = − 1

4T

∫ ∫f (w, z)dw dz + ψ(w) + φ(z).

13. Find the general solution of Eq. (∗) in Exercise 12 in terms of x and t, ifF(x, t) = T cos(t).

3.4 One-Dimensional Wave Equation: Generalities 233

3.4 One-Dimensional Wave Equation: Generalities

As for the one-dimensional heat equation, we can make some comments fora generalized one-dimensional wave equation. For the sake of generality, weassume that some nonuniform properties are present. For the sake of simplic-ity, we assume that the equation is homogeneous and free of u. Our initialvalue–boundary value problem will be

∂

∂x

(s(x)

∂u

∂x

)= p(x)

c2

∂2u

∂t2, l < x < r, 0 < t, (1)

α1u(l, t) − α2∂u

∂x(l, t) = c1, 0 < t, (2)

β1u(r, t) + β2∂u

∂x(r, t) = c2, 0 < t, (3)

u(x,0) = f (x), l < x < r, (4)

∂u

∂t(x,0) = g(x), l < x < r. (5)

We assume that the functions s(x) and p(x) are positive for l ≤ x ≤ r, be-cause they represent physical properties, that s, s′, and p are all continuous,and that s and p have no dimensions. Also, suppose that none of the coeffi-cients α1, α2, β1, β2 are negative.

To obtain homogeneous boundary conditions we can write

u(x, t) = v(x) + w(x, t),

just as before. In the wave equation, however, neither of the names “steady-state solution” nor “transient solution” is appropriate; for, as we shall see, thereis no steady state, or limiting case, nor is there a part of the solution that tendsto zero as t tends to infinity. Nevertheless, v represents an equilibrium solu-tion, and, more important, it is a useful mathematical device to consider u inthe form provided in the preceding equation.

The function v(x) is required to satisfy the conditions

(sv′)′ = 0, l < x < r,

α1v(l) − α2v′(l) = c1,

β1v(r) + β2v′(r) = c2.

Thus v(x) is exactly equivalent to the “steady-state solution” discussed for theheat equation.


The function w(x, t), being the difference between u(x, t) and v(x), satisfiesthe initial value–boundary value problem

∂

∂x

(s(x)

∂w

∂x

)= p(x)

c2

∂2w

∂t2, l < x < r, 0 < t, (6)

α1w(l, t) − α2∂w

∂x(l, t) = 0, 0 < t, (7)

β1w(r, t) + β2∂w

∂x(r, t) = 0, 0 < t, (8)

w(x,0) = f (x) − v(x), l < x < r, (9)

∂w

∂t(x,0) = g(x), l < x < r. (10)

Since the equation and the boundary conditions are homogeneous and lin-ear, we attempt a solution by separation of variables. If w(x, t) = φ(x)T(t), wefind in the usual way that the factor functions φ and T must satisfy

T ′′ + c2λ2T = 0, 0 < t, (11)(s(x)φ′)′ + λ2p(x)φ = 0, l < x < r, (12)

α1φ(l) − α2φ′(l) = 0, (13)

β1φ(r) + β2φ′(r) = 0. (14)

The eigenvalue problem represented in the last three lines is a regularSturm–Liouville problem, because of the assumptions we have made abouts, p, and the coefficients. We know that there are an infinite number of non-negative eigenvalues λ2

1, λ22, . . . and corresponding eigenfunctions φ1, φ2, . . .

that have the orthogonality property∫ r

lφn(x)φm(x)p(x)dx = 0, n �= m.

The solution of the equation for T is

Tn(t) = an cos(λnct) + bn sin(λnct).

From here it is clear that the frequencies of vibration that occur in the solutionof Eqs. (1)–(3) are λnc/2π (cycles per time unit). Thus, it is the eigenvaluescoming from Eqs. (12)–(14) that determine these frequencies.

Having solved the subsidiary problems that arose after separation of vari-ables, we can begin to assemble the solution. The function w will have theform

w(x, t) =∞∑

n=1

φn(x)(an cos(λnct) + bn sin(λnct)

), (15)

3.4 One-Dimensional Wave Equation: Generalities 235

and its two initial conditions, yet to be satisfied, are

w(x,0) =∞∑

n=1

anφn(x) = f (x) − v(x), l < x < r,

∂w

∂t(x,0) =

∞∑n=1

bnλncφn(x) = g(x), l < x < r.

By employing the orthogonality of the φn, we determine that the coefficientsan and bn are given by

an = 1

In

∫ r

l

[f (x) − v(x)

]φn(x)p(x)dx, (16)

bn = 1

Inλnc

∫ r

lg(x)φn(x)p(x)dx, (17)

where

In =∫ r

lφ2

n(x)p(x)dx. (18)

Finally, u(x, t) = v(x) + w(x, t) is the solution of the original problem, andeach of its parts is completely specified. From the form of w(x, t), we can makecertain observations about u.

1. u(x, t) does not have a limit as t → ∞. Each term of the series form of w

is periodic in time and thus does not die away.

2. Except in very special cases, the eigenvalues λ2n are not closely related to

each other. So in general, if u causes acoustic vibrations, the result willnot be musical to the ear. (A sound would be musical if, for instance,λn = nλ1, as in the case of the uniform string.)

3. In general, u(x, t) is not even periodic in time. Although each term in theseries for w is periodic, the terms do not have a common period (except inspecial cases), and so the sum is not periodic.

E X E R C I S E S

1. Verify the formulas for the an and bn. Under what conditions on f and gcan we say that the initial conditions are satisfied?

2. Check the statement that v(x) is the same for the heat conduction problemand for the problem considered here.

3. Identify the period of Tn(t) and the associated frequency.


4. Although u(x, t) has no limit as t → ∞, show that the following general-ized limit is valid:

v(x) = limT→∞

1

T

∫ T

0u(x, t)dt.

(Hint: Do the integration and limiting term by term.)

5. Formally solve the problem

∂

∂x

(s(x)

∂u

∂x

)= p(x)

c2

(∂2u

∂t2+ γ

∂u

∂t

)+ q(x)u, l < x < r, 0 < t,

with the boundary conditions Eqs. (2) and (3) and initial conditionsEqs. (4) and (5), taking γ to be constant.

6. Verify that w(x, t) as given in Eq. (15) satisfies the differential equation andthe boundary conditions Eqs. (6)–(8).

7. In reference to the observations at the end of the section, prove the follow-ing statement: The product solutions of the problem in Eqs. (6)–(8) all havea common period in time if the eigenvalues of the problem in Eqs. (12)–(14) satisfy the relation

λn = α(n + β),

where β is a rational number.

8. Find a separation-of-variables solution of the problem

∂u2

∂x2= 1

c2

(∂2u

∂t2+ γ 2u

), 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = 0, 0 < t,

u(x, t) = f (x),∂u

∂t(x,0) = g(x), 0 < x < a.

Is this an instance of the problem in this section? Which of the observationsat the end of the section are valid for the solution of this problem?

3.5 Estimation of EigenvaluesIn many instances, one is interested not in the full solution to the wave equa-tion, but only in the possible frequencies of vibration that may occur. For ex-ample, it is of great importance that bridges, airplane wings, and other struc-tures not vibrate; so it is important to know the frequencies at which a struc-ture can vibrate, in order to avoid them. By inspecting the solution of the gen-eralized wave equation, which we investigate in the preceding section, we can

3.5 Estimation of Eigenvalues 237

see that the frequencies of vibration are λnc/2π , n = 1,2,3, . . . . Thus we mustfind the eigenvalues λ2

n in order to identify the frequencies of vibration.Consider the following Sturm–Liouville problem:(

s(x)φ′)′ − q(x)φ + λ2p(x)φ = 0, l < x < r, (1)

φ(l) = 0, φ(r) = 0, (2)

where s, s′,q, and p are continuous and s and p are positive for l ≤ x ≤ r. (Notethat we have a rather general differential equation but very special boundaryconditions.)

If φ1 is the eigenfunction corresponding to the smallest eigenvalue λ21, then

φ1 satisfies Eq. (1) for λ = λ1. Alternatively, we can write

−(sφ′

1

)′ + qφ1 = λ21pφ1, l < x < r.

Multiplying through this equation by φ1 and integrating from l to r, we obtain∫ r

l−(

sφ′1

)′φ1 dx +

∫ r

lqφ2

1 dx = λ21

∫ r

lpφ2

1 dx.

If the first integral is integrated by parts, it becomes

−sφ′1φ1

∣∣r

l+

∫ r

lsφ′

1φ′1 dx.

But φ1(l) = φ1(r) = 0, so the first term vanishes and we are left with the equal-ity ∫ r

ls[φ′

1

]2dx +

∫ r

lqφ2

1 dx = λ21

∫ r

lpφ2

1 dx.

Because p(x) is positive for l ≤ x ≤ r, the integral on the right is positive andwe may define λ2

1 as

λ21 =

∫ rl s[φ′

1]2 dx + ∫ rl qφ2

1 dx∫ rl pφ2

1 dx= N(φ1)

D(φ1). (3)

It can be shown that, if y(x) is any function with two continuous derivatives(l ≤ x ≤ r) that satisfies y(l) = y(r) = 0, then

λ21 ≤ N(y)

D(y). (4)

By choosing any convenient function y that satisfies the boundary conditions,we obtain from the ratio N(y)/D(y) an upper bound on λ2

1. Usually this boundis quite a good estimate. One should keep in mind that the graph of the eigen-function φ1(x) does not cross the x-axis between l and r, so the graph of y(x)should not cross the axis either.


Example.Estimate the first eigenvalue of

φ′′ + λ2φ = 0, 0 < x < 1,

φ(0) = φ(1) = 0.

Let us try y(x) = x(1 − x), which satisfies the boundary conditions. Theny′(x) = 1 − 2x and

N(y) =∫ 1

0

[y′(x)

]2dx =

∫ 1

0(1 − 2x)2 dx = 1

3,

D(y) =∫ 1

0y2(x)dx =

∫ 1

0x2(1 − x)2 dx = 1

30.

Therefore, N(y)/D(y) = 10. We know, of course, that φ1(x) = sin(πx), and

N(φ1) =∫ 1

0π2 cos2(πx)dx = π2

2, D(φ1) =

∫ 1

0sin2(πx)dx = 1

2,

so N(φ1)/D(φ1) = λ21 = π2 < 10, confirming Eq. (4). �

Example.Estimate the first eigenvalue of

(xφ′)′ + λ2 1

xφ = 0, 1 < x < 2,

φ(1) = φ(2) = 0.

The integrals to be calculated are

N(y) =∫ 2

1x(y′)2 dx, D(y) =

∫ 2

1

1

xy2 dx.

The tabulation gives results for several trial functions. It is known that the firsteigenvalue and eigenfunction are

λ21 =

(π

ln 2

)2 ∼= 20.5423,

φ1(x) = sin

(π ln x

ln 2

).

The error for the best of the trial functions is about 1.44%

y(x)√

x(2 − x)(x − 1) (2 − x)(x − 1)(2 − x)(x − 1)

xN(y)

D(y)23.7500 22.1349 20.8379

�

3.6 Wave Equation in Unbounded Regions 239

This method of estimating the first eigenvalue is called Rayleigh’s method,and the ratio N(y)/D(y) is called the Rayleigh quotient. In some mechanicalsystems, the Rayleigh quotient may be interpreted as the ratio between poten-tial and kinetic energy. There are many other methods for estimating eigenval-ues and for systematically improving the estimates.

E X E R C I S E S

1. Using Eq. (3), show that if q ≥ 0, then λ21 ≥ 0 also.

2. Verify the results for at least one of the trial functions used in the secondexample.

3. Estimate the first eigenvalue of the problem

φ′′ + λ2(1 + x)φ = 0, 0 < x < 1,

φ(0) = φ(1) = 0.

4. Verify that the general solution of the following differential equation isax cos(λ/x) + bx sin(λ/x), and then solve the eigenvalue problem.

φ′′ + λ2

x4φ = 0, 1 < x < 2,

φ(1) = 0, φ(2) = 0.

5. Estimate the lowest eigenvalue of the problem in Exercise 4 using y =(x − 1)(2 − x).

6. Estimate the lowest eigenvalue of the problem

φ′′ + λ2xφ = 0, 0 < x < 1,

φ(0) = 0, φ(1) = 0.

Use the trial function xb(1 − x), and minimize the Rayleigh quotient withrespect to b.

3.6 Wave Equation in Unbounded RegionsWhen the wave equation is to be solved for 0 < x < ∞ or for −∞ <

x < ∞, we can proceed as we did for the solution of the heat equation in theseunbounded regions. That is to say, we separate variables and use a Fourierintegral to combine the product solutions.


Consider the problem

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < t, 0 < x, (1)

u(x,0) = f (x), 0 < x, (2)

∂u

∂t(x,0) = g(x), 0 < x, (3)

u(0, t) = 0, 0 < t. (4)

We require in addition that the solution u(x, t) be bounded as x → ∞.On separating variables, we make u(x, t) = φ(x)T(t) and find that the fac-

tors satisfy

T ′′ + λ2c2T = 0, 0 < t,

φ′′ + λ2φ = 0, 0 < x,

φ(0) = 0, |φ(x)| bounded.

The solutions are easily found to be

φ(x;λ) = sin(λx), T(t;λ) = A cos(λct) + B sin(λct),

and we combine the products φ(x;λ)T(t;λ) in a Fourier integral

u(x, t) =∫ ∞

0

(A(λ) cos(λct) + B(λ) sin(λct)

)sin(λx)dλ. (5)

The initial conditions become

u(x,0) = f (x) =∫ ∞

0A(λ) sin(λx)dλ, 0 < x,

∂u

∂t(x,0) = g(x) =

∫ ∞

0λcB(λ) sin(λx)dλ, 0 < x.

Both of these equations are Fourier integrals. Thus the coefficient functionsare given by

A(λ) = 2

π

∫ ∞

0f (x) sin(λx)dx, B(λ) = 2

πλc

∫ ∞

0g(x) sin(λx)dx.

It is sufficient to demand that∫ ∞

0 |f (x)|dx and∫ ∞

0 |g(x)|dx both be finite inorder to guarantee the existence of A and B.

The deficiency of the Fourier integral form of the solution given in Eq. (5)is that the formula gives no idea of what u(x, t) looks like. The d’Alembert


solution of the wave equation can come to our aid again here. We know thatthe solution of Eq. (1) has the form

u(x, t) = ψ(x + ct) + φ(x − ct).

The two initial conditions boil down to

ψ(x) + φ(x) = f (x), 0 < x,

ψ(x) − φ(x) = G(x) + A, 0 < x.

As in the finite case, we have defined

G(x) = 1

c

∫ x

0g(y)dy

and A is any constant.From the two initial conditions we obtain

ψ(x) = 1

2

(f (x) + G(x) + A

), x > 0,

φ(x) = 1

2

(f (x) − G(x) − A

), x > 0.

Both f and G are known for x > 0. Thus

ψ(x + ct) = 1

2

(f (x + ct) + G(x + ct) + A

)is defined for all x > 0 and t ≥ 0. But φ(x− ct) is not yet defined for x− ct < 0.That means that we must extend the functions f and G in such a way as todefine φ for negative arguments and also satisfy the boundary condition. Thesole boundary condition is Eq. (4), which becomes

u(0, t) = 0 = ψ(ct) + φ(−ct).

In terms of f and G, extensions of f and G, this is

0 = f (ct) + G(ct) + A + f (−ct) − G(−ct) − A.

Since f and G are not dependent on each other, we must have individually

f (ct) + f (−ct) = 0, G(ct) − G(−ct) = 0.

That is, f is fo, the odd extension of f , and G is Ge, the even extension of G.Finally, we arrive at a formula for the solution:

u(x, t) = 1

2

[fo(x + ct) + Ge(x + ct)

] + 1

2

[fo(x − ct) − Ge(x − ct)

]. (6)


Figure 5 Solution of Eqs. (1)–(4) with g(x) ≡ 0. On the left are graphs offo(x + ct) (solid) and of fo(x − ct) (dashed) at the times shown. On the right arethe graphs of u(x, t) for 0 < x, made by averaging the graphs on the left.

Now, given the functions f (x) and g(x), it is a simple matter to construct fo

and Ge and thus to graph u(x, t) as a function of either variable or to evaluate itfor specific values of x and t. By way of illustration, Fig. 5 shows the solution ofEqs. (1)–(4) as a function of x at various times, for f (x) as shown and g(x) ≡ 0.

Another interesting problem that can be treated by the d’Alembert methodis one in which the boundary condition is a function of time. For simplicity,we take zero initial conditions. Our problem becomes

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < t, 0 < x, (7)

u(x,0) = 0, 0 < x, (8)

∂u

∂t(x,0) = 0, 0 < x, (9)


u(0, t) = h(t), 0 < t. (10)

As u is to be a solution of the wave equation, it must have the form

u(x, t) = ψ(x + ct) + φ(x − ct). (11)

The two initial conditions, Eqs. (8) and (9), can be treated exactly as in thefirst problem. Of course, G(x) ≡ 0, and the constant A, being arbitrary, maybe taken as 0. The conclusion is that

ψ(x) = 0, φ(x) = 0, 0 < x.

Because both x and t are positive in this problem, we see that ψ(x + ct) = 0always, so Eq. (11) may be simplified to

u(x, t) = φ(x − ct). (12)

The boundary condition Eq. (10) will tell us how to evaluate φ for negativearguments. The equation is

u(0, t) = φ(−ct) = h(t), 0 < t. (13)

We now put together what we know of the function φ:

φ(q) =

0, q > 0,

h

(−q

c

), q < 0.

(14)

The argument q is a dummy, used to avoid association with either x or t. Equa-tions (12) and (14) now specify the solution u(x, t) completely.

Example.Take h(t) as shown in Fig. 6. The major steps to construct the graph of φ(q).Note that the graph of φ for negative argument is that of h, reflected. In otherwords, to make the graph of φ(q), start from the graph of h(t): (1) Graph theeven extension he(t); (2) replace the right half (from 0 up) with 0; (3) adjustscales so that q = −c where t = −1, etc.

The graphs in Fig. 7 show u(x, t) as a function of x for various values of t. Itis clear from both the graphs and the formula that the disturbance caused bythe variable boundary condition arrives at a fixed point x at time x/c. Thus thedisturbance travels with the velocity c, the wave speed. An example is animatedon the CD. �

A wave equation accompanied by nonzero initial conditions and time-varying boundary conditions can be solved by breaking it into two prob-lems, one like Eqs. (1)–(4) with zero boundary condition, and the other likeEqs. (7)–(10) with zero initial conditions.


Figure 6 Graphs of h(t) and φ(q) for semi-infinite string with time-varyingboundary condition.

Figure 7 Graphs of u(x, t) versus x for the semi-infinite string under thetime-varying boundary condition u(0, t) = h(t), with h(t) as shown in Fig. 6. Theshape seen in the last drawing will continue to travel to the right.


E X E R C I S E S

1. Derive a formula similar to Eq. (6) for the case in which the boundary con-dition Eq. (4) is replaced by

∂u

∂x(0, t) = 0, 0 < t.

2. Derive Eq. (6) from Eq. (5) by using trigonometric identities for the prod-uct sin(λx) · cos(λct), and so forth, and recognizing certain Fourier inte-grals.

3. Sketch the solution of Eqs. (1)–(4) as a function of x at times t = 0, 1/2c,1/c, 2/c, 3/c, if g(x) = 0 everywhere and f (x) is the rectangular pulse

f (x) ={0, 0 < x < 1,

1, 1 < x < 2,0, 2 < x.

4. Same as Exercise 3, but f (x) = 0 and g(x) is the rectangular pulse

g(x) ={0, 0 < x < 1,

c, 1 < x < 2,0, 2 < x.

5. Sketch the solution of Eqs. (7)–(10) at times t = 0, π/2, π , 3π/2, 2π , 5π/2,if h(t) = sin(t). Take c = 1.

6. Sketch the solution of Eqs. (7)–(10) at times t = 0, 1/2, 3/2, and 5/2, ifc = 1 and

h(t) ={0, 0 < t < 1,

1, 1 < t < 2,0, 2 < t.

7. Use the d’Alembert solution of the wave equation to solve the problem

∂2u

∂x2= 1

c2

∂2u

∂t2, −∞ < x < ∞, 0 < t,

u(x,0) = f (x), −∞ < x < ∞,

∂u

∂t(x,0) = g(x), −∞ < x < ∞.

8. The solution of the problem stated in Exercise 7 is sometimes written

u(x, t) = 1

2

(f (x + ct) + f (x − ct)

) + 1

2c

∫ x+ct

x−ctg(z)dz.


Show that this is correct. You will need Leibniz’s rule (see the Appendix) todifferentiate the integral.


The wave equation is one of the oldest equations of mathematical physics.Euler, Bernoulli, and d’Alembert all solved the problem of the vibrating stringabout 1750, using either separation of variables or what we called d’Alembert’smethod. This latter is, in fact, a very special case of the method of characteris-tics, in essence a way of identifying new independent variables having specialsignificance. Street’s Analysis and Solution of Partial Differential Equations hasa chapter on characteristics, including their use in numerical solutions. Wan’sMathematical Models and Their Analysis gives applications to traffic flow andalso discusses other wave phenomena. (See the Bibliography.)

Because many physical phenomena described by the wave equation arepart of our everyday experience — the sounds of musical instruments, forinstance — they are often featured in popular expositions of mathematicalphysics. The book of Davis and Hersch (The Mathematical Experience) ex-plains standing waves (product solutions) and superposition in an elemen-tary way. Of course, many other phenomena are described by the wave equa-tion. Among the most important for modern life are electrical and mag-netic waves, which are solutions of special cases of the Maxwell field equa-tions. These and other kinds of waves (including water waves) are studied inMain’s Vibrations and Waves in Physics; both exposition and figures are firstrate.

The potential difference V between the interior and exterior of a nerve axoncan be modeled approximately by the Fitzhugh–Nagumo equations,

∂V

∂t= ∂2V

∂x2+ V − 1

3V3 − R,

∂R

∂t= k(V + a − bR).

Here R represents a restoring effect and a, b, and k are constants. At first glance,one would expect V to behave like the solution of a heat equation. But a trav-eling wave solution,

V(x, t) = F(x − ct), R(x, t) = G(x − ct),

of these equations can be found that shows many important features of nerveimpulses. This system and many other exciting biological applications ofmathematics are reported by Murray’s excellent book, Mathematical Biology.


More information about the Rayleigh quotient and estimation of eigenval-ues is in Boundary and Eigenvalue Problems in Mathematical Physics, by Sagan.The classic reference for eigenvalues, and indeed for the partial differentialequations of mathematical physics in general, is the work by Courant andHilbert, Methods of Mathematical Physics.

Chapter ReviewSee the CD for Review Questions.

Miscellaneous ExercisesExercises 1–5 refer to the problem

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = 0, 0 < t,

u(x,0) = f (x),∂u

∂t(x,0) = g(x), 0 < x < a.

1. Take f (x) = 1, 0 < x < a, and g(x) ≡ 0. (This is rather unrealis-tic if u(x, t) is the displacement of a vibrating string.) Find a series(separation-of-variables) solution.

2. Sketch u(x, t) of Exercise 1 as a function of x at various times throughoutone period.

3. The solution u(x, t) of Exercise 1 takes on only the three values 1, 0, and−1. Make a sketch of the region 0 < x < a, 0 < t, and locate the placeswhere u takes on each of the values.

4. Take g(x) = 0 and f (x) to be this function:

f (x) =

3hx

2a, 0 < x <

2a

3,

3h(a − x)

a,

2a

3< x < a.

The graph of f is triangular, with peak at x = 2a/3. Find a series solutionfor u(x, t).

5. Sketch u(x, t) of Exercise 4 as a function of x at times 0 to a/c in steps ofa/6c.


6. Find an analytic (integral) solution of this wave problem

∂2u

∂x2= 1

c2

∂2u

∂t2, −∞ < x < ∞, 0 < t,

u(x,0) = f (x),∂u

∂t(x,0) = g(x), −∞ < x < ∞,

with g(x) = 0 and

f (x) ={

h, |x| < ε,

0, |x| > ε.

7. Sketch the solution of the problem in Exercise 6 at times t = 0, ε/c, 2ε/c,3ε/c.

8. Same as Exercise 7, but f (x) = 0 and

g(x) ={

c, |x| < ε,

0, |x| > ε.

9. Let u(x, t) be the solution of the problem

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x, 0 < t,

u(0, t) = 0, 0 < t,

u(x,0) = f (x),∂u

∂t(x,0) = g(x), 0 < x.

Sketch the solution u(x, t) as a function of x at times t = 0, a/6c, a/2c,5a/6c, 7a/6c. Use g(x) ≡ 0 and

f (x) =

3hx

2a, 0 < x <

2a

3,

3h(a − x)

a,

2a

3< x < a,

0, a < x.

10. Same task as in Exercise 9 but

f (x) ={

sin(x), 0 < x < π ,

0, π < x

and g(x) = 0. Sketch at times t = 0, π/4c, π/2c, 3π/4c, π/c, 2π/c.

11. Let u(x, t) be the solution of the wave equation on the semi-infinite in-terval 0 < x < ∞, with both initial conditions equal to zero but with the


time-varying boundary condition

u(0, t) =

sin

(ct

a

), 0 < t <

πa

c,

0,πa

c< t.

Sketch u(x, t) as a function of x at various times.

12. Same as Exercise 11, but the boundary condition is u(0, t) = h, for allt > 0.

13. Same as Exercise 11, but the boundary condition is

u(0, t) =

hct

a, 0 < t <

a

c,

h(2a − ct)

a,

a

c< t <

2a

c,

0,2a

c< t.


(eαxφ′)′ + λ2eαxφ = 0, 0 < x < 1,

φ(0) = 0, φ(1) = 0.

(This problem can be solved exactly.)


φ′′ − xφ + λ2φ = 0, 0 < x < 1,

φ(0) = 0, φ(1) = 0.

16. Show that the nonlinear wave equation

∂u

∂t+ u

∂u

∂x+ ∂3u

∂x3= 0

(the Korteweg–deVries equation) has, as one solution,

u(x, t) = 12a2sech2(ax − 4a3t).

A wave of this form is called a soliton or solitary wave.

17. The solution in Exercise 16 is of the form u(x, t) = f (x − ct). What is thefunction f , and what is the wave speed c?


18. For t < 0, water flows steadily through a long pipe connected at x = 0to a large reservoir and open at x = a to the air. At time t = 0, a valve atx = a is suddenly closed. Reasonable expressions for the conservation ofmomentum and of mass are

∂u

∂t= −∂p

∂x, 0 < x < a, 0 < t, (A)

∂p

∂t= −c2 ∂u

∂x, 0 < x < a, all t, (B)

where p is gauge pressure and u is mass flow rate. If the pipe is rigid,c2 = K/ρ, the ratio of bulk modulus of water to its density. Show thatboth p and u satisfy the wave equation. The phenomenon described hereis called water hammer.

19. Introduce a function v with the definition u = ∂v/∂x, p = −∂v/∂t.Show that (A) becomes an identity and that (B) becomes the wave equa-tion for v.

20. Reasonable boundary and initial conditions for u and p are

u(x,0) = U0 (constant), 0 < x < a,

p(x,0) = 0, 0 < x < a,

p(0, t) = 0, all t,

u(a, t) = 0, t > 0.

Restate these as conditions on v. Show that the first and third equationsmay be replaced by

v(x,0) = U0x, 0 < x < a,

v(0, t) = 0, all t.

21. Solve the problem in Exercise 20; find a series form for v(x, t).

22. In many problems involving fluid flow, the combination

∂u

∂t+ V

∂u

∂x

(called the Stokes derivative) appears. Here V is the speed of the fluid inthe x-direction. If V equals u or otherwise depends on u, this operator isnonlinear and difficult to work with. Let us assume that V is a constant,so that the operator is linear, and define new variables

ξ = x + Vt, τ = x − Vt, u(x, t) = v(ξ, τ ).


Show that

∂u

∂t+ V

∂u

∂x= 2V

∂v

∂ξ.

23. Assume that u(x, y, t) has the product form shown in what follows. Sep-arate the variables in the given partial differential equation.

u(x, y, t) = ψ(x + Vt)φ(x − Vt)Y(y),

∂2u

∂y2= 1

k

(∂u

∂t+ V

∂u

∂x

).

24. A fluid flows between two parallel plates held at temperature 0. At theinlet, fluid temperature is T0 and initially the fluid is at temperature T1.If V is the speed of the fluid in the x-direction, a problem describing thetemperature u(x, y, t) is

∂2u

∂y2= 1

k

(∂u

∂t+ V

∂u

∂x

), 0 < y < b, 0 < x, 0 < t,

u(x,0, t) = 0, u(x,b, t) = 0, 0 < x, 0 < t,

u(0, y, t) = T0, 0 < y < b, 0 < t,

u(x, y,0) = T1, 0 < x, 0 < y < b.

Make a separation of variables as in Exercise 23. State and solve the eigen-value problem for Y . Show that

un(x, y, t) = φn(x − Vt) exp(−λ2

nk(x + Vt)/2V)

sin(λny)

satisfies the partial differential equation and boundary conditions at y =0 and y = b, without restriction of φn (except differentiability).

25. Show how to satisfy the initial and inlet conditions in the problem of Ex-ercise 24, by forming a sum of product solutions and correctly choosingthe φn.

26. Find all functions φ such that u(x, t) = φ(x − ct) is a solution of the heatequation,

∂2u

∂x2= 1

k

∂u

∂t.

27. Take the constant c = (1 + i)√

ωk/2 in Exercise 26 and show that thefunctions

e−px sin(ωt − px), e−px cos(ωt − px)


can be obtained from φ(x − ct) and φ(x − ct). (Here p = √ω/2k and c is

the complex conjugate of c. Refer to Exercise 6 in Chapter 2, Section 10.)

28. Some nonlinear equations can also result in “traveling wave solutions,”u(x, t) = φ(x − ct). Show that Fisher’s equation,

∂2u

∂x2= ∂u

∂t− u(1 − u),

has a solution of this form if φ satisfies the nonlinear differential equa-tion

φ′′ + cφ′ + φ(1 − φ) = 0.

29. Show that the function u is a solution of the problem

∂2u

∂x2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t,

u(0, t) = 0, u(a, t) = sin(ωt), 0 < t,

provided that the parameters are such that sin(ωa/c) �= 0.

u(x, t) = sin(ωx/c) sin(ωt)

sin(ωa/c).

30. If ω = πc/a, the denominator of the function in Exercise 29 is 0. Showthat, for this value of ω, a function satisfying the wave equation and thegiven boundary condition is

u(x, t) = − ct

asin

(πx

a

)cos

(πct

a

)− x

acos

(πx

a

)sin

(πct

a

).

31. A string in a musical instrument is typically not as flexible as assumedin Section 1. For such a string, the displacement u may satisfy the partialdifferential equation

∂2u

∂x2− ε

∂4u

∂x4= 1

c2

∂2u

∂t2, 0 < x < a, 0 < t,

where c2 = T/ρ, as in Section 1, and ε = EI/T, with E = Young’s mod-ulus for the material, I = second moment of area. (See an elasticity ref-erence.)

Assuming that u(x, t) = φ(x)T(t), carry out a separation of variableand find the eigenvalue problem for φ. Take the boundary conditions tobe


u(0, t) = 0, u(a, t) = 0,

∂2u

∂x2(0, t) = 0,

∂2u

∂x2(a, t) = 0, 0 < t.

32. Show that φ(x) = sin(µx) satisfies the eigenvalue problem found in Ex-ercise 31, provided that µ = nπ/a and

λ2 = µ2 + εµ4,

where −λ2 = T ′′(t)/c2T(t).

33. The values of λ2 are related to the frequencies of vibration of the stringmentioned in Exercise 31. Show that λn approaches nπ/a for any fixed nas ε approaches 0.

34. The longitudinal vibration of a thin rod has been described by Love (seeBibliography) with the equation

∂2u

∂x2= 1

c2

(∂2u

∂t2− ε

∂4u

∂x2∂t2

), 0 < x < a, 0 < t.

Here, u(x, t) is the displacement of the points that are at x when thereis no motion; ε = νK2, where ν is Poisson’s ratio and K is the radius ofgyration of a cross section of the rod. Take boundary conditions

u(0, t) = 0, u(a, t) = 0,

and separate the variables by assuming u(x, t) = φ(x)T(t). It will be use-ful to name φ′′/φ = −λ2.

35. The eigenvalue problem in Exercise 34 is routine. Once that is solved,find the differential equation for T(t), solve it, and determine the fre-quencies of vibration.

The PotentialEquation C H A P T E R

4

4.1 Potential EquationThe equation for the steady-state temperature distribution in two dimensions(see Chapter 5) is

∂2u

∂x2+ ∂2u

∂y2= 0.

The same equation describes the equilibrium (time-independent) displace-ments of a two-dimensional membrane, and so is an important common partof both the heat and wave equations in two dimensions. Many other physicalphenomena — gravitational and electrostatic potentials, certain fluid flows —and an important class of functions are described by this equation, thus mak-ing it one of the most important of mathematics, physics, and engineering.The analogous equation in three dimensions is

∂2u

∂x2+ ∂2u

∂y2+ ∂2u

∂z2= 0.

Either equation may be written ∇2u = 0 and is commonly called the potentialequation or Laplace’s equation.

The solutions of the potential equation (called harmonic functions) havemany interesting properties. An important one, which can be understood in-tuitively, is the maximum principle: If ∇2u = 0 in a region, then u cannothave a relative maximum or minimum inside the region unless u is constant.(Thus, if ∂u/∂x and ∂u/∂y are both zero at some point, it is a saddle point.)If u is thought of as the steady-state temperature distribution in a metal plate,

255

256 Chapter 4 The Potential Equation

it is clear that the temperature cannot be greater at one point than at all othernearby points. For if such were the case, heat would flow away from the hotpoint to cooler points nearby, thus reducing the temperature at the hot point.But then the temperature would not be unchanging with time. We return tothis matter in Section 4.

A complete boundary value problem consists of the potential equation in aregion plus boundary conditions. These may be of any of the three types

u given,∂u

∂ngiven, or αu + β

∂u

∂ngiven

along any section of the boundary. (By ∂u/∂n, we mean the directional deriv-ative in the direction normal, or perpendicular, to the boundary.) When u isspecified along the whole boundary, the problem is called Dirichlet’s problem;if ∂u/∂n is specified along the whole boundary, it is Neumann’s problem. Thesolutions of Neumann’s problem are not unique, for if u is a solution, so is uplus a constant.

It is often useful to consider the potential equation in other coordinate sys-tems. One of the most important is the polar coordinate system, in which thevariables are

r = √x2 + y2, θ = tan−1

(y

x

),

x = r cos(θ), y = r sin(θ).

By convention we require r ≥ 0. We shall define

u(x, y) = u(r cos(θ), r sin(θ)

) = v(r, θ)

and find an expression for the Laplacian of u,

∇2u = ∂2u

∂x2+ ∂2u

∂y2,

in terms of v and its derivatives by using the chain rule. The calculations areelementary but tedious. (See Exercise 7.) The results are

∂2u

∂x2= cos2(θ)

∂2v

∂r2− 2 sin(θ) cos(θ)

r

∂2v

∂θ∂r+ sin2(θ)

r2

∂2v

∂θ2

+ sin2(θ)

r

∂v

∂r+ 2 sin(θ) sin(θ)

r2

∂v

∂θ,

∂2u

∂y2= sin2(θ)

∂2v

∂r2+ 2 sin(θ) cos(θ)

r

∂2v

∂θ∂r+ cos2(θ)

r2

∂2v

∂θ2

+ cos2(θ)

r

∂v

∂r− 2 sin(θ) sin(θ)

r2

∂v

∂θ.

Chapter 4 The Potential Equation 257

From these equations we easily find that the Laplacian in polar coordinates is

∇2v = ∂2v

∂r2+ 1

r

∂v

∂r+ 1

r2

∂2v

∂θ2= 1

r

∂

∂r

(r∂v

∂r

)+ 1

r2

∂2v

∂θ2.

In cylindrical (r, θ, z) coordinates, the Laplacian is

∇2v = 1

r

∂

∂r

(r∂v

∂r

)+ 1

r2

∂2v

∂θ2+ ∂2v

∂z2.

E X E R C I S E S

1. Find a relation among the coefficients of the polynomial

p(x, y) = a + bx + cy + dx2 + exy + fy2

that makes it satisfy the potential equation. Choose a specific polynomialthat satisfies the equation, and show that, if ∂p/∂x and ∂p/∂y are both zeroat some point, the surface there is saddle shaped.

2. Show that u(x, y) = x2 −y2 and u(x, y) = xy are solutions of Laplace’s equa-tion. Sketch the surfaces z = u(x, y). What boundary conditions do thesefunctions fulfill on the lines x = 0, x = a, y = 0, y = b?

3. If a solution of the potential equation in the square 0 < x < 1, 0 < y < 1has the form u(x, y) = Y(y)sin(πx), of what form is the function Y? Find afunction Y that makes u(x, y) satisfy the boundary conditions u(x,0) = 0,u(x,1) = sin(πx).

4. Find a function u(x), independent of y, that satisfies the potential equation.

5. What functions v(r), independent of θ , satisfy the potential equation inpolar coordinates?

6. Show that rn sin(nθ) and rn cos(nθ) both satisfy the potential equation inpolar coordinates (n = 0,1,2, . . .).

7. Find expressions for the partial derivatives of u with respect to x and y interms of derivatives of v with respect to r and θ .

8. If u and v are the x- and y-components of the velocity in a fluid, it can beshown (under certain assumptions) that these functions satisfy the equa-tions

∂u

∂x+ ∂v

∂y= 0, (A)

∂u

∂x− ∂v

∂x= 0. (B)


(a) (b)

(c) (d)

Figure 1 (a) u is displacement of a membrane; the graph of f (x) is an isoscelestriangle. (b) u is the temperature on a cross section of a long bar. (c) u is voltagein a rectangular sheet of conducting material. (d) φ is a velocity potential (seeExercise 8). What are x- and y-velocities on the boundaries?

Show that the definition of a velocity potential function φ by the equations

u = −∂φ

∂x, v = −∂φ

∂y

causes (B) to be identically satisfied and turns (A) into the potential equa-tion. (See Section 4.7, Comments and References, at the end of this chap-ter.)

9. For each of the diagrams in Fig. 1, (a) write out the problem in mathemat-ical form (partial differential equation and boundary conditions); (b) pro-vide an interpretation in words of the boundary conditions for the giveninterpretation of the unknown function.

4.2 Potential in a Rectangle 259

4.2 Potential in a RectangleOne of the simplest and most important problems in mathematical physics isDirichlet’s problem in a rectangle. To take an easy case, we consider a problemwith just two nonzero boundary conditions:

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, 0 < y < b, (1)

u(x,0) = f1(x), 0 < x < a, (2)

u(x,b) = f2(x), 0 < x < a, (3)

u(0, y) = 0, 0 < y < b, (4)

u(a, y) = 0, 0 < y < b. (5)

It is not immediately clear that separation of variables will work. However,we have a homogeneous partial differential equation and some homogeneousboundary conditions, so we can try the method. If we assume that u(x, y) hasa product form u = X(x)Y(y), then Eq. (1) becomes

X′′(x)Y(y) + X(x)Y ′′(y) = 0.

This equation can be separated by dividing through by XY to yield

X′′(x)

X(x)= −Y ′′(y)

Y(y). (6)

The nonhomogeneous conditions Eqs. (2) and (3) will not, in general, becomeconditions on X or Y , but the homogeneous conditions Eqs. (4) and (5), asusual, require that

X(0) = 0, X(a) = 0. (7)

Now, both sides of Eq. (6) must be constant, but the sign of the constantis not obvious. If we try a positive constant (say, µ2), Eq. (6) represents twoordinary equations:

X′′ − µ2X = 0, Y ′′ + µ2Y = 0.

The solutions of these equations are

X(x) = A cosh(µx) + B sinh(µx), Y(y) = C cos(µy) + D sin(µy).

In order to make X satisfy the boundary conditions Eq. (7), both A and B mustbe zero, leading to a solution u(x, y) ≡ 0. Thus we try the other possibility forsign, taking both members in Eq. (6) to equal −λ2.


Under the new assumption, Eq. (6) separates into

X′′ + λ2X = 0, Y ′′ − λ2Y = 0. (8)

The first of these equations, along with the boundary conditions, is recogniz-able as an eigenvalue problem, whose solutions are

Xn(x) = sin(λnx), λ2n =

(nπ

a

)2

.

The functions Y that accompany the X’s are

Yn(y) = an cosh(λny) + bn sinh(λny).

The a’s and b’s are for the moment unknown.We see that Xn(x)Yn(y) is a solution of the (homogeneous) potential Eq. (1),

which satisfies the homogeneous conditions Eqs. (4) and (5). A sum of thesefunctions should satisfy the same conditions and equation, so u may have theform

u(x, y) =∞∑

n=1

(an cosh(λny) + bn sinh(λny)

)sin(λnx). (9)

The nonhomogeneous boundary conditions Eqs. (2) and (3) are yet to besatisfied. If u is to be of the form of Eq. (9), the boundary condition Eq. (2)becomes

u(x,0) =∞∑

n=1

an sin

(nπx

a

)= f1(x), 0 < x < a. (10)

We recognize a problem in Fourier series immediately. The an must be theFourier sine coefficients of f1(x),

an = 2

a

∫ a

0f1(x) sin

(nπx

a

)dx.

The second boundary condition reads

u(x,b) =∞∑

n=1

(an cosh(λnb) + bn sinh(λnb)

)sin

(nπx

a

)

= f2(x), 0 < x < a.

This also is a problem in Fourier series, but it is not as neat. The constant

an cosh(λnb) + bn sinh(λnb)


must be the nth Fourier sine coefficient of f2. Since an is known, bn can bedetermined from the following computations:

an cosh(λnb) + bn sinh(λnb) = 2

a

∫ a

0f2(x) sin(λnx)dx = cn,

bn = cn − an cosh(λnb)

sinh(λnb).

If we use this last expression for bn and substitute into Eq. (9), we find thesolution

u(x, y) =∞∑

n=1

{cn

sinh(λny)

sinh(λnb)

+ an

[cosh(λny) − cosh(λnb)

sinh(λnb)sinh(λny)

]}sin(λnx). (11)

Notice that the function multiplying cn is 0 at y = 0 and is 1 at y = b. Similarly,the function multiplying an is 1 at y = 0 and 0 at y = b. An easier way to writethis latter function is

sinh(λn(b − y))

sinh(λnb),

as can readily be found from hyperbolic identities.

Example.Suppose f1 and f2 are both given by

f1(x) = f2(x) =

2x

a, 0 < x <

a

2,

2

(a − x

a

),

a

2< x < a.

Then

cn = an = 8

π2

sin(nπ/2)

n2.

The solution of the potential equation for these boundary conditions is

u(x, y) = 8

π2

∞∑n=1

sin(

nπ2

)n2

sinh(

nπa y

) + sinh(

nπa (b − y)

)sinh

(nπb

a

) sin

(nπx

a

). (12)

In Fig. 2 is a graph of some level curves, u(x, y) = constant, for the case wherea = b, and also a view of the surface z = u(x, y). Also see color figures on theCD. �


(a) (b)

Figure 2 (a) Level curves of the solution u(x, y) of the example problem (seeEq. (12)) for the case b = a = 1. Each curve is part of the locus of points thatsatisfy u(x, y) = constant for constants 0 to 0.9 in steps of 0.1. For some constants,the locus consists of more than one connected curve. (b) Perspective view of thesurface z = u(x, y).

Now we have seen a solution of Dirichlet’s problem in a rectangle with ho-mogeneous conditions on two parallel sides. In general, of course, the bound-ary conditions will be nonhomogeneous on all four sides of the rectangle. Butthis more general problem can be broken down into two problems like the onewe have solved.

Consider the problem

∇2u = 0, 0 < x < a, 0 < y < b, (13)

u(x,0) = f1(x), 0 < x < a, (14)

u(x,b) = f2(x), 0 < x < a, (15)

u(0, y) = g1(y), 0 < y < b, (16)

u(a, y) = g2(y), 0 < y < b. (17)

Let u(x, y) = u1(x, y) + u2(x, y). We will put conditions on u1 and u2 so thatthey can readily be found, and from them u can be put together. The mostobvious conditions are the following:

∇2u1 = 0, ∇2u2 = 0,

u1(x,0) = f1(x), u2(x,0) = 0,

u1(x,b) = f2(x), u2(x,b) = 0,

u1(0, y) = 0, u2(0, y) = g1(y),

u1(a, y) = 0, u2(a, y) = g2(y).


It is evident that u1 + u2 is the solution of the original problem Eqs. (13)–(17). Also, each of the functions u1 and u2 has homogeneous conditions onparallel boundaries. We already have determined the form of u1. The otherfunction would be of the form

u2(x, y) =∞∑

n=1

sin(µny)An sinh(µnx) + Bn sinh(µn(a − x))

sinh(µna), (18)

where µn = nπ/b and

An = 2

b

∫ b

0g2(y) sin(µny)dy,

Bn = 2

b

∫ b

0g1(y) sin(µny)dy.

In the individual problems for u1 and u2, the technique of separation ofvariable works because the homogeneous conditions on parallel sides of therectangle can be translated into conditions on one of the factor functions.

When the boundary conditions are not complicated functions, it may bepossible to satisfy some of them with a polynomial function. (See Exercises 1and 2 of Section 4.1.) Then the difference between u and the polynomial is asolution of the potential equation that satisfies some homogeneous boundaryconditions.

E X E R C I S E S

1. Show that sinh(λy) and sinh(λ(b − y)) are independent solutions ofY ′′ − λ2Y = 0 with λ �= 0. Thus a combination of these two functions mayreplace a combination of sinh and cosh as the general solution of this dif-ferential equation.

2. Show that the solution of the example problem may be written

u(x, y) = 8

π2

∞∑n=1

sin(

nπ2

)n2

cosh(

nπa (y − 1

2 b))

cosh(

nπb2a

) sin

(nπx

a

).

3. Use the form in Exercise 2 to compute u in the center of the rectangle inthe three cases b = a, b = 2a, b = a/2. (Hint: Check the magnitude of theterms.)

4. Verify that each term of Eq. (9) satisfies Eqs. (1), (4), and (5).


5. Solve the problem

∇2u = 0, 0 < x < a, 0 < y < b,

u(0, y) = 0, u(a, y) = 0, 0 < y < b,

u(x,0) = 0, u(x,b) = f (x), 0 < x < a,

where f is the same as in the example. Sketch some level curves of u(x, y).

6. Solve the potential problem on the rectangle 0 < x < a, 0 < y < b, subjectto the boundary conditions u(a, y) = 1, 0 < y < b, and u = 0 on the rest ofthe boundary.

7. Solve the problem of the potential equation in the rectangle 0 < x < a, 0 <

y < b, for each of the following sets of boundary conditions. Before solving,make a pictorial version of the problem as in Exercise 9 of Section 4.1.

a. u(x,b) = 100, 0 < x < a; u = 0 on the other three sides of the rectangle.

b. u(x,b) = 100, 0 < x < a; u(a, y) = 100, 0 < y < b; u = 0 on the othertwo sides of the rectangle.

c. u(x,b) = bx, 0 < x < a; u(a, y) = ay, 0 < y < b; u = 0 on the other twosides of the rectangle.

8. Solve the problem for u2. (That is, derive Eq. (18).)

4.3 Further Examples for a RectangleIn Section 4.2, we solved Dirichlet problems with separation of variables. Thesame method applies to problems with other types of boundary conditions, asshown in the following.

Example 1.In this problem, the unknown function might be a voltage in a conductor. Theleft and right sides are electrically insulated.

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, 0 < y < b,

∂u

∂x(0, y) = 0,

∂u

∂x(a, y) = 0, 0 < y < b,

u(x,0) = 0, u(x,b) = V0x/a, 0 < x < a.

We have homogeneous conditions on the facing sides at x = 0 and x = a. Ifwe look for solutions in the product form u(x, y) = X(x)Y(y), we find (as ex-

4.3 Further Examples for a Rectangle 265

pected) that

X′′(x)

X(x)= −Y ′′(y)

Y(y)= constant.

The conditions at x = 0 and x = a become

X′(0) = 0, X′(a) = 0.

If we make the separation constant −λ2, we find a familiar eigenvalue problemfor X whose solution is

X0(x) = 1, λ0 = 0,

Xn(x) = cos(λnx), λn = nπ/a, n = 1,2, . . . .

For the factor Y(y), the differential equation is

Y ′′0 = 0, or Y ′′

n − λ2nYn = 0

with solution

Y0(y) = a0 + b0y or Yn(y) = an cosh(λny) + bn sinh(λny).

Thus, the principle of superposition leads to the series solution

u(x, y) = a0 + b0y +∞∑

n=1

(an cosh(λny) + bn sinh(λny)

)cos(λnx).

The boundary condition at y = 0 becomes

a0 +∞∑1

an cos(λnx) = 0, 0 < x < a,

from which we see that all the a’s are 0. Then at y = b we have

b0b +∞∑

n=1

(bn sinh(λnb)

)cos(λnx) = V0x

a, 0 < x < a.

This is a slightly disguised cosine series. The coefficients are

b0b = 1

a

∫ a

0V0

(x

a

)dx,

bn sinh(λnb) = 2

a

∫ a

0V0

(x

a

)cos(λnx)dx.

See a color graphic of the solution on the CD. �


We have seen that the success of the separation of variables method dependson having homogeneous boundary conditions at the ends of one of the inter-vals involved. In Section 4.2 we mentioned splitting up a Dirichlet problem,if necessary, to achieve this. The same splitting technique applies in problemswhere boundary condition of other kinds are used. The principle is to zeroconditions on two facing sides of the region and to copy the rest.

Example 2.This problem may describe the temperature u(x, y) in a thin plate betweeninsulating sheets.

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, 0 < y < b,

∂u

∂x(0, y) = 0, u(a, y) = Sy, 0 < y < b,

∂u

∂y(x,0) = S, u(x,b) = Sbx

a, 0 < x < a.

Since we have nonhomogeneous conditions on adjacent sides, we must splitthe problem in order to solve by separation of variables. Here are the two prob-lems:

∂2u1

∂x2+ ∂2u1

∂y2= 0,

∂2u2

∂x2+ ∂2u2

∂y2= 0,

∂u1

∂x(0, y) = 0, u1(a, y) = 0,

∂u2

∂x(0, y) = 0, u2(a, y) = Sy,

∂u1

∂y(x,0) = S, u1(x,b) = Sbx

a,

∂u2

∂y(x,0) = 0, u2(x,b) = 0.

The solution of the original problem is the sum u = u1 + u2. Here is thereasoning in detail.

1. The potential equation is linear and homogeneous. By the Principle ofSuperposition, the sum of solutions is a solution.

2. At x = a we have u(a, y) = 0 + Sy, and at y = b we have u(x,b) =Sbx/a + 0. Both conditions are satisfied.

3. From elementary calculus, we know

∂u

∂x= ∂u1

∂x+ ∂u2

∂x,

∂u

∂y= ∂u1

∂y+ ∂u2

∂y.


Then at the left and bottom boundaries, we have

∂u

∂x(0, y) = 0 + 0,

∂u

∂y(x,0) = S + 0.

These are satisfied as well.

Thus, it remains to solve the two problems for u1 and u2. (See the Exercises.)Here are product solutions. For u1:

cos(λnx)(an cosh(λny) + bn sinh(λny)

), λn =

(n − 1

2

)π

a, n = 1,2, . . . .

For u2:

cos(µny)(An cosh(µnx) + Bn sinh(µnx)

), µn =

(n − 1

2

)π

b, n = 1,2, . . . .

�

The simple polynomial solutions that we found in Section 4.1, Exercise 1,can be very useful in reducing the number of series needed for a solution. Ifnonhomogeneous conditions are given on adjacent sides and these are con-stants or first-degree polynomials in one variable, then a polynomial may beable to satisfy enough of them to simplify the work.

Example 3.Refer to the problem in Example 2. The polynomial v(y) = Sy satisfies thepotential equation and several of the boundary conditions:

∂v

∂x(0, y) = 0, v(a, y) = Sy, 0 < y < b,

∂v

∂y(x,0) = S, v(x,b) = Sb, 0 < x < a.

Thus, we may set u(x, y) = v(y) + w(x, y) and determine that w must be thesolution of this problem, similar to the problem for u2 in Example 2:

∂2w

∂x2+ ∂2w

∂y2= 0, 0 < x < a, 0 < y < b,

∂w

∂x(0, y) = 0, w(a, y) = 0, 0 < y < b,

∂w

∂y(x,0) = 0, w(x,b) = Sb(x − a)

a, 0 < x < a.

The solution is left as an exercise. �


Poisson EquationMany problems in engineering and physics require the solution of the Poissonequation,

∇2u = −H in a region R.

Here are three examples of such problems.

(1) u is the deflection of a membrane that is fastened at its edges, so u = 0 onthe boundary of R; H is proportional to the pressure difference acrossthe membrane. (See Section 5.1.)

(2) u is the steady-state temperature in a cross section of a long cylindricalrod that is carrying an electrical current; H is proportional to the powerin resistance heating. (See Section 5.2.)

(3) u is the stress function on the cross section R of a cylindrical bar or rodin torsion (the shear stresses are proportional to the partial derivativesof u); H is proportional to the rate of twist and to the shear modulus ofthe material; u = 0 on the boundary of R.

If H is a constant, a polynomial of the form

P(x, y) = A + Bx + Cy + Dx2 + Exy + Fy2

is a solution of Poisson’s equation, provided that

2(D + F) = −H.

The other coefficients are arbitrary and may be chosen for convenience in sat-isfying boundary conditions.

Example 4.Find the deflection u of a membrane that is modeled by this problem. Theconstant is H = p/σ , where p is the pressure difference (below to above) andσ is the surface tension in the membrane.

A polynomial can be chosen that satisfies the partial differential equationand boundary conditions on facing sides. For instance,

v(x) = Hx(a − x)

2

satisfies the Poisson equation and two boundary conditions,

v(0) = 0, v(a) = 0.


Thus, we may set u(x, y) = v(x) + w(x, y) and determine that w is a solutionof the problem

∂2w

∂x2+ ∂2w

∂y2= 0, 0 < x < a, 0 < y < b,

w(0, y) = 0, w(a, y) = 0, 0 < y < b,

w(x,0) = −v(x), w(x,b) = −v(x), 0 < x < a.

The CD has color graphics of the solution. �

In general, if H is a polynomial in x and y, a solution can be found in theform of a polynomial of total degree 2 higher than H. If H is a more gen-eral function, it may be expressed as a double Fourier series (see Chapter 5),and the partial differential equation can be solved following the idea of Sec-tion 1.11B.

E X E R C I S E S

1. Solve the problem consisting of the potential equation on the rectangle0 < x < a, 0 < y < b with the given boundary conditions. Two of the threeare very easy if a polynomial is subtracted from u.

a.∂u

∂x(0, y) = 0; u = 1 on the remainder of the boundary.

b.∂u

∂x(0, y) = 0,

∂u

∂x(a, y) = 0; u(x,0) = 0, u(x,b) = 1.

c.∂u

∂x(x,0) = 0, u(x,b) = 0; u(0, y) = 1, u(a, y) = 0.

2. Same task as Exercise 1.

a. u(x,b) = 100; the outward normal derivative is 0 on the rest of theboundary.

b. u(x,b) = 100, u(0, y) = 0, u(a, y) = 100,∂u

∂y(x,0) = 0.

3. Finish the work for Example 1: Find the bn, form the series, and check thatall conditions are satisfied.

4. In Example 2, check that the given product solution for u1(x, y) satisfiesthe conditions and determine the coefficients an and bn.

5. In Example 2, check that the given product solution for u2(x, y) satisfiesthe conditions and determine the coefficients An and Bn.


6. Explain the difference between the cosine series in Example 1 and the co-sine series for u1(x, y) in Example 2. What is the source of the difference?

7. Finish the work for Example 3. That is, find w(x, y) as a series and checkthat the boundary conditions are all satisfied.

8. Compare the amount of work involved in solving the problem of Exam-ple 2 (including Exercises 4 and 5) with the work for Example 3 (includingExercise 7).

9. Finish the work of Example 4: Find the solution, as a series, for w(x, y).Form u(x, y) and use the first term of the series for w(x, y) to obtain anexpression for the value of u( a

2 ,b2 ). This would be the maximum deflection

of the membrane.

10. Find the condition on the coefficients so that the following generalsecond-degree polynomial is a solution of the Poisson equation, ∇2p =−H, where H is constant:

p(x, y) = A + Bx + Cy + Dx2 + Exy + Fy2.

11. Same task as Exercise 10, but H = K(x2 + y2) and p is this part of thegeneral fourth-degree polynomial

p(x, y) = Ax4 + Bx3y + Cx2y2 + Dxy3 + Ey4,

∂2u

∂x2+ ∂2u

∂y2= H, 0 < x < a, 0 < y < b,

u(0, y) = 0, u(a, y) = 0, 0 < x < a,

u(x,0) = 0, u(x,b) = 0, 0 < y < b.

4.4 Potential in Unbounded RegionsThe potential equation, as well as the heat and wave equations, can be solvedin unbounded regions. Consider the following problem, in which the regioninvolved is half a vertical strip, or a slot:

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, 0 < y, (1)

u(x,0) = f (x), 0 < x < a, (2)

u(0, y) = g1(y), 0 < y, (3)

u(a, y) = g2(y), 0 < y. (4)

As usual, we required that u(x, y) remain bounded as y → ∞.

4.4 Potential in Unbounded Regions 271

In order to make the separation of variables work, we must break thisup into two problems. Following the model of Sections 4.2 and 4.3 we setu(x, y) = u1(x, y) + u2(x, y) and require that the parts satisfy these two solv-able problems:

∇2u1 = 0, ∇2u2 = 0, 0 < x < a, 0 < y,

u1(x,0) = f (x), u2(x,0) = 0, 0 < x < a,

u1(0, y) = 0, u2(0, y) = g1(y), 0 < y,

u1(a, y) = 0, u2(a, y) = g2(y), 0 < y.

We attack the problem for u1 by assuming the product form and separatingvariables:

u1(x, y) = X(x)Y(y),X′′(x)

X(x)= −Y ′′(y)

Y(y)= −λ2.

The sign of the constant −λ2 is determined by the boundary conditions atx = 0 and x = a, which become homogeneous conditions on the factor X(x):

X(0) = 0, X(a) = 0. (5)

(We also can see that the condition to be satisfied along y = 0 demands func-tions of x that permit a representation of an arbitrary function.)

The boundary conditions, Eq. (5), together with the differential equation

X′′ + λ2X = 0 (6)

that comes from the separation of variables, constitute a familiar eigenvalueproblem, whose solution is

Xn(x) = sin

(nπx

a

), λ2

n =(

nπ

a

)2

, n = 1,2,3, . . . .

The equation for Y is

Y ′′ − λ2Y = 0, 0 < y.

In addition to satisfying this differential equation, Y must remain bounded asy → ∞. The solutions of the equation are eλy and e−λy. Of these, the first isunbounded, so

Yn(y) = exp(−λny).

Finally, we can write the solution of the first problem as

u1(x, y) =∞∑

n=1

an sin

(nπx

a

)exp

(−nπy

a

). (7)

The constants an are to be determined from the condition at y = 0.


The solution of the second problem is somewhat different. Again we seeksolutions in the product form u2(x, y) = X(x)Y(y). The homogeneous bound-ary condition at y = 0 and the boundedness condition become conditions onY(y):

Y(0) = 0, Y(y) bounded as y → ∞.

Then the potential equation becomes

X′′(x)

X(x)+ Y ′′(y)

Y(y)= 0, (8)

and both ratios must be constant. If Y ′′/Y is positive, the auxiliary conditionsforce Y to be identically 0. Thus, we take Y ′′/Y = −µ2, or Y ′′ + µ2Y = 0, andfind that the solution that satisfies the auxiliary conditions is

Y(y) = sin(µy),

for any µ > 0. Then the general solution of the equation X′′/X = µ2 is

X(x) = Asinh(µx)

sinh(µa)+ B

sinh(µ(a − x))

sinh(µa).

We have chosen this special form on the basis of our experience in solving thepotential equation in the rectangle.

Since µ is a continuous parameter, we combine our product solutions bymeans of an integral, finding

u2(x, y) =∫ ∞

0

[A(µ)

sinh(µx)

sinh(µa)+ B(µ)

sinh(µ(a − x))

sinh(µa)

]sin(µy)dµ. (9)

The nonhomogeneous boundary conditions at x = 0 and x = a are satisfied if

u2(0, y) =∫ ∞

0B(µ) sin(µy)dµ = g1(y), 0 < y,

u2(a, y) =∫ ∞

0A(µ) sin(µy)dµ = g2(y), 0 < y.

Obviously these two equations are Fourier integral problems, so we know howto determine the coefficients A(µ) and B(µ). An example of this kind of prob-lem is shown on the CD.

The potential equation can also be solved in a strip (0 < x < a, −∞ < y< ∞), a quarter-plane (0 < x, 0 < y), or a half-plane (0 < x, −∞ < y < ∞).Along each boundary line, a boundary condition is imposed, and the solutionis required to remain bounded in remote portions of the region considered. Ingeneral, a Fourier integral is employed in the solution, because the separationconstant is a continuous parameter, as in the second problem here.

4.4 Potential in Unbounded Regions 273

E X E R C I S E S

1. Find a formula for the constants an in Eq. (7).

2. Verify that u1(x, y) in the form given in Eq. (7) satisfies the potential equa-tion and the homogeneous boundary conditions.

3. Find formulas for A(µ) and B(µ) of Eq. (9).

4. Solve the potential equation in the slot, 0 < x < a, 0 < y, for each of thesesets of boundary conditions.

a. u(0, y) = 0, u(a, y) = 0, 0 < y; u(x,0) = 1, 0 < x < a;

b. u(0, y) = 0, u(a, y) = e−y, 0 < y; u(x,0) = 0, 0 < x < a;

c. u(0, y) = f (y) ={

1, 0 < y < b,0, b < y,

u(a, y) = 0, 0 < y;

u(x,0) = 0, 0 < x < a.

5. Solve the potential equation in the slot, 0 < x < a, 0 < y, for each of thesesets of boundary conditions.

a.∂u

∂x(0, y) = 0, u(a, y) = 0, 0 < y; u(x,0) = 1, 0 < x < a;

b.∂u

∂x(0, y) = 0, u(a, y) = e−y, 0 < y;

u(x,0) = 0, 0 < x < a;

c. u(0, y) = 0, u(a, y) = f (y) ={

1, 0 < y < b,0, b < y;

∂u

∂y(x,0) = 0, 0 < x < a.

6. Show that if the separation constant had been chosen as −µ2 instead ofµ2 in solving for u2 (leading to Y ′′ − µ2Y = 0), then Y(y) ≡ 0 is theonly function that satisfies the differential equation, satisfies the conditionY(0) = 0, and remains bounded as y → ∞.

7. Solve the problem of potential in a slot under the boundary conditions

u(x,0) = 1, u(0, y) = u(a, y) = e−y.

8. Show that the function v(x, y) given here satisfies the potential equationand the boundary conditions on the “long” sides in Exercise 7, provided


that cos(a/2) �= 0:

v(x, y) = cos(x − 1

2 a)

cos(

12 a

) e−y.

What partial differential equation and boundary conditions are satisfiedby w(x, y) = u(x, y) − v(x, y) if u is the function of Exercise 7?

9. Solve the potential equation in the slot 0 < y < b, 0 < x for each of thefollowing sets of boundary conditions:

a. u(0, y) = 0, u(x,0) = 0, u(x,b) = f (x) ={1, 0 < x < a,

0, a < x;

b. u(0, y) = 0, u(x,0) = e−x, u(x,b) = 0.

10. Find product solutions of this potential problem in a strip:

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, −∞ < y < ∞,

subject to the boundedness condition u(x, y) bounded as y → ±∞.

11. Solve the potential problem consisting of the equation and boundednessconditions from Exercise 10 and the boundary conditions

u(0, y) = 0, u(a, y) = e−|y|, −∞ < y < ∞.

12. Show how to solve the potential problem of Exercise 10 together with theboundary conditions

u(0, y) = g1(y), u(a, y) = g2(y), −∞ < y < ∞,

where g1 and g2 are suitable functions.

13. Find product solutions of this potential problem in the quarter-plane:

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x, 0 < y,

u(0, y) = 0, u(x,0) = f (x).

Note that u(x, y) must remain bounded as x → ∞ and as y → ∞.

14. Solve the potential equation in the quarter-plane, x > 0, y > 0, subject tothe boundary conditions

u(0, y) = e−y, y > 0; u(x,0) = e−x, x > 0.

4.5 Potential in a Disk 275

15. Find product solutions of the potential equation in the half-plane y > 0:

∂2u

∂x2+ ∂2u

∂y2= 0, −∞ < x < ∞, 0 < y < ∞,

u(x,0) = f (x), −∞ < x < ∞.

What boundedness conditions must u(x, y) satisfy?

16. Convert your solution of Exercise 15 into the following formula (see Ex-ercise 8 of Section 4.4 and Section 2.11):

u(x, y) = 1

π

∫ ∞

−∞f (x′)

y

y2 + (x − x′)2dx′.

17. Use the formula in Exercise 16 to solve the potential problem in the upperhalf-plane, with boundary condition

u(x,0) = f (x) ={1, 0 < x,

0, x < 0.

18. Solve the problem stated in Exercise 15 if the boundary function is

f (x) ={

1, |x| < a,0, |x| > a.

19. Show that u(x, y) = x is the solution of the potential equation in a slotunder the boundary conditions f (x) = x, g1(y) = 0, g2(y) = a. Can thissolution be found by the method of this section?

4.5 Potential in a DiskIf we need to solve the potential equation in a circular disk x2 + y2 < c2, it isnatural to use polar coordinates r, θ , in terms of which the disk is describedby 0 < r < c. We found in Section 4.1 that the potential equation in polarcoordinates is

1

r

∂

∂r

(r∂v

∂r

)+ 1

r2

∂2v

∂θ2= 0.

There are some special features of this coordinate system. First, it is clear thatsome coefficients of the Laplacian are negative powers of r. Thus, we mustenforce a boundedness condition at r = 0. Second, θ and θ + 2π refer to thesame angle. Therefore, we must require that the function v(r, θ) be periodicwith period 2π in θ .


The Dirichlet problem on a disk can now be stated as

1

r

∂

∂r

(r∂v

∂r

)+ 1

r2

∂2v

∂θ2= 0, 0 ≤ r < c, (1)

v(c, θ) = f (θ), (2)

v(r, θ + 2π) = v(r, θ), 0 < r < c, (3)

v(r, θ) bounded as r → 0+. (4)

By assuming v(r, θ) = R(r)Q(θ) we can separate variables. The potential equa-tion becomes

1

r

(rR′)′

Q + 1

r2RQ′′ = 0.

Separation is effected by dividing through by RQ/r2:

r(rR′(r))′

R(r)+ Q′′(θ)

Q(θ)= 0.

As usual, both terms must be constant. We know that if Q′′/Q is a posi-tive constant, then Q will be exponential, not periodic. Therefore we chooseQ′′/Q = −λ2 and obtain this singular eigenvalue problem:

Q′′ + λ2Q = 0, (5)

Q(θ + 2π) = Q(θ). (6)

The accompanying equation for R(r) is

r(rR′)′ − λ2R = 0, (7)

R(r) bounded as r → 0+. (8)

The general solution of Eq. (5) (if λ > 0) is

Q(θ) = A cos(λθ) + B sin(λθ).

This function is periodic for all λ, but the period is 2π only if λ is an inte-ger. Thus we have λn = n, n = 1,2, . . . . In addition, if λ = 0, we have a peri-odic solution that is any constant. Thus, the solution of the singular eigenvalueproblem of Eqs. (5) and (6) is

λ0 = 0, Q0(θ) = 1,

λn = n, Qn(θ) = A cos(nθ) + B sin(nθ), n = 1,2,3, . . . .

The novelty here is that we have two eigenfunctions for each eigenvalue n =1,2, . . . .


Knowing that λ2n = n2, we can easily find R(r). The equation for R becomes

r2R′′ + rR′ − n2R = 0, 0 < r < c,

when the indicated differentiations are carried out. This is a Cauchy–Eulerequation, whose solutions are known to have the form R(r) = rα , where α

is constant. Substituting R = rα , R′ = αrα−1, and R′′ = α(α − 1)rα−2 into itleaves (

α(α − 1) + α − n2)rα = 0, 0 < r < c.

Because rα is not zero, the constant factor in parentheses must be zero — thatis, α = ±n. The general solution of the differential equation is any combina-tion of rn and r−n. The latter, however, is unbounded as r approaches zero,so we discard that solution, retaining Rn(r) = rn. In the special case n = 0, thetwo solutions are the constant function 1 and ln(r). The logarithm is discardedbecause of its behavior at r = 0.

Now we reassemble our solution. The functions

r0 · 1 = 1, rn cos(nθ), rn sin(nθ) (9)

are all solutions of the potential equation, so a general linear combination ofthese solutions will also be a solution. Thus v(r, θ) may have the form

v(r, θ) = a0 +∞∑

n=1

anrn cos(nθ) +∞∑

n=1

bnrn sin(nθ). (10)

At the true boundary r = c, the boundary condition reads

v(c, θ) = a0 +∞∑

n=1

cn(an cos(nθ) + bn sin(nθ)

) = f (θ), −π < θ ≤ π.

This is a Fourier series problem, as in Section 1.1, solved by choosing

a0 = 1

2π

∫ π

−π

f (θ)dθ,

an = 1

πcn

∫ π

−π

f (θ) cos(nθ)dθ, bn = 1

πcn

∫ π

−π

f (θ) sin(nθ)dθ. (11)

Example.Consider the problem consisting of Eqs. (1)–(4) with

v(c, θ) = f (θ) ={0, −π < θ < −π/2,

1, −π/2 < θ < π/2,0, π/2 < θ < π .


The solution is given by Eq. (10), provided that the coefficients are chosenaccording to Eq. (11). Since f (θ) is an even function, bn = 0, and

a0 = 1

π

∫ π

0f (θ)dθ = 1

2,

an = 2

πcn

∫ π

0f (θ) cos(nθ)dθ = 2 sin(nπ/2)

nπcn.

Therefore, the solution of the problem is

v(r, θ) = 1

2+

∞∑n=1

2 sin(nπ/2)

nπ

rn

cncos(nθ). (12)

The level curves of this function are all arcs of circles that pass through theboundary points r = c, θ = ±π/2, where f (θ) jumps between 0 and 1. Alongthe x-axis, the function has the simple closed form 1/2 + (2/π) tan−1(x/c).

The CD has a color graphic of the solution. �

Properties of the SolutionNow that we have the form Eq. (10) of the solution of the potential equation,we can see some important properties of the function v(r, θ). In particular, bysetting r = 0 we obtain

v(0, θ) = a0 = 1

2π

∫ π

−π

f (θ)dθ = 1

2π

∫ π

−π

v(c, θ)dθ.

This says that the solution of the potential equation at the center of a disk isequal to the average of its values around the edge of the disk. It is easy to showalso that

v(0, θ) = 1

2π

∫ π

−π

v(r, θ)dθ (13)

for any r between 0 and c! This characteristic of solutions of the potential equa-tion is called the mean value property. From the mean value property, it is justa step to prove the maximum principle mentioned in Section 4.1, for the meanvalue of a function lies between the minimum and the maximum and cannotequal either unless the function is constant.

An important consequence of the maximum principle — and thus of themean value property — is a proof of the uniqueness of the solution of theDirichlet problem. Suppose that u and v are two solutions of the potentialequation in some region R and that they have the same values on the bound-ary of R. Then their difference, w = u − v, is also a solution of the potentialequation in R and has value 0 all along the boundary of R. By the maximum


principle, w has maximum and minimum values 0, and therefore w is identi-cally 0 throughout R. In other words, u and v are identical.

E X E R C I S E S

1. Solve the potential equation in the disk 0 < r < c if the boundary condi-tion is v(c, θ) = |θ |, −π < θ ≤ π .

2. Same as Exercise 1 if v(c, θ) = θ , −π < θ < π . Is the boundary conditionsatisfied at θ = ±π ?

3. Same as Exercise 1, with boundary condition

v(c, θ) = f (θ) ={

cos(θ), −π/2 < θ < π/2,0, otherwise.

4. Find the value of the solution at r = 0 for the problems of Exercises 1, 2,and 3.

5. If the function f (θ) in Eq. (2) is continuous and sectionally smooth andsatisfies f (−π+) = f (π−), what can be said about convergence of theseries for v(c, θ)?

6. Show that

v(r, θ) = a0 +∞∑

n=1

r−n(an cos(nθ) + bn sin(nθ)

)

is a solution of Laplace’s equation in the region r > c (exterior of a disk)and has the property that |v(r, θ)| is bounded as r → ∞.

7. If the condition v(c, θ) = f (θ) is given, what are the formulas for the a’sand b’s in Exercise 6?

8. The solution of Eqs. (1)–(4) can be written in a single formula by thefollowing sequence of operations:

a. Replace θ by φ in Eq. (11) for the a’s and b’s;

b. replace the a’s and b’s in Eq. (10) by the integrals in part a;

c. use the trigonometric identity

cos(nθ) cos(nφ) + sin(nθ) sin(nφ) = cos(n(θ − φ)

);d. take the integral outside the series;


e. add up the series (see Section 1.10, Exercise 5a). Then v(r, θ) is givenby the single integral (Poisson integral formula)

v(r, θ) = 1

2π

∫ π

−π

f (φ)c2 − r2

c2 + r2 − 2rc cos(θ − φ)dφ.

9. Solve Laplace’s equation in the quarter-disk 0 < θ < π/2, 0 < r < c, sub-ject to the boundary conditions v(r,0) = 0, v(r,π/2) = 0, v(c, θ) = 1.

10. Generalize the results of Exercise 9 by solving this problem:

1

r

∂

∂r

(r∂v

∂r

)+ 1

r2

∂2v

∂θ2= 0, 0 < θ < απ, 0 < r < c,

v(r,0) = 0, v(r, απ) = 0, 0 < r < c,

v(c, θ) = f (θ), 0 < θ < απ.

Here, α is a parameter between 0 and 2.

11. Suppose that α > 1 in Exercise 10. Show that there is a product solutionwith the property that ∂v

∂r (r, θ) is not bounded as r → 0+.

4.6 Classification of Partial Differential Equationsand Limitations of the Product Method

By this time, we have seen a variety of equations and solutions. We have con-centrated on three different, homogeneous equations (heat, wave, and poten-tial) and have found the qualitative features summarized in the following table:

Equation FeaturesHeat Exponential behavior in time. Existence of a limiting (steady-state) solution.

Smooth graph for t > 0.

Wave Oscillatory (not always periodic) behavior in time. Retention of discontinuitiesfor t > 0.

Potential Smooth surface. Maximum principle. Mean value property.

These three two-variable equations are the most important representativesof the three classes of second-order linear partial differential equations in twovariables. The most general equation that fits this description is

A∂2u

∂ξ 2+ B

∂2u

∂ξ∂η+ C

∂2u

∂η2+ D

∂u

∂ξ+ E

∂u

∂η+ Fu + G = 0,

4.6 Classification and Limitations 281

where A,B,C, and so forth are, in general, functions of ξ and η. (We use Greekletters for the independent variables to avoid implying any relations to spaceor time.) Such an equation can be classified according to the sign of B2 − 4AC:

B2 − 4AC < 0: elliptic,

B2 − 4AC = 0: parabolic,

B2 − 4AC > 0: hyperbolic.

Because A,B, and C are functions of ξ and η (not of u), the classification of anequation may vary from point to point. It is easy to see that the heat equationis parabolic, the wave equation is hyperbolic, and the potential equation iselliptic. The classification of an equation determines important features of thesolution and also dictates the method of attack when numerical techniques areused for solution.

The question naturally arises whether separation of variables works on allequations. The answer is no. For instance, the equation

(ξ + η2

)∂2u

∂ξ 2+ ∂2u

∂η2= 0

does not admit separation of variables. In general, it is difficult to say justwhich equations can be solved by this method. However, it is necessary to haveB ≡ 0.

The region in which the solution is to be found also limits the applicabilityof the method we have used. The region must be a generalized rectangle. Bythis we mean a region bounded by coordinate curves of the coordinate systemof the partial differential equation. Put another way, the region is describedby inequalities on the coordinates, whose endpoints are fixed quantities. Forinstance, we have worked in regions described by the following sets of inequal-ities:

0 < x < a, 0 < t,

0 < x, 0 < t,

−∞ < x < ∞, 0 < t,

0 < x < a, 0 < y < b,

0 < r < c, −π < θ ≤ π.

All of these are generalized rectangles, but only one is an ordinary rectangle.An L-shaped region is not a generalized rectangle, and our methods wouldbreak down if applied to, for instance, the potential equation there.

There are, as we know, restrictions on the kinds of boundary conditionsthat can be handled. From the examples in this chapter it is clear that we need


homogeneous or “homogeneous-like” conditions on opposite sides of a gener-alized rectangle. Examples of “homogeneous-like” conditions are the require-ment that a function remain bounded as some variable tends to infinity, or theperiodic conditions at θ = ±π (see Section 4.5). The point is that if two ormore functions satisfy the conditions, so does a sum of those functions.

In spite of the limitations of the method of separation of variables, it workswell on many important problems in two or more variables and provides in-sight into the nature of their solutions. Moreover, it is known that in thosecases where separation of variables can be carried out, it will find a solution ifone exists.

E X E R C I S E S

1. Classify the following equations.

a.∂2u

∂x∂y= 0;

b.∂2u

∂x2+ ∂2u

∂x∂y+ ∂2u

∂y2= 2x;

c.∂2u

∂x2− ∂2u

∂x∂y+ ∂2u

∂y2= 2u;

d.∂2u

∂x2− 2

∂2u

∂x∂y+ ∂2u

∂y2= ∂u

∂y;

e.∂2u

∂x2− ∂2u

∂y2− ∂u

∂y= 0.

2. Show that, in polar coordinates, an annulus, a sector, and a sector of anannulus are all generalized rectangles.

3. In which of the equations in Exercise 1 can the variables be separated?

4. Sketch the regions listed in the text as generalized rectangles.

5. Solve these three problems and compare the solutions.

a.∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < 1, 0 < y,

u(x,0) = f (x), 0 < x < 1,

u(0, y) = 0, u(1, y) = 0, 0 < y;

b.∂2u

∂x2= ∂2u

∂y2, 0 < x < 1, 0 < y,


u(x,0) = f (x),∂u

∂y(x,0) = 0, 0 < x < 1,

u(0, y) = 0, u(1, y) = 0, 0 < y;

c.∂2u

∂x2= ∂u

∂y, 0 < x < 1, 0 < y,

u(x,0) = f (x), 0 < x < 1,

u(0, y) = 0, u(1, y) = 0, 0 < y.

6. Show that if f1, f2, . . . all satisfy the periodic boundary conditions

f (−π) = f (π), f ′(−π) = f ′(π),

then so does the function c1f1 + c2f2 + · · · , where the c’s are constants.

7. Longitudinal waves in a slender rod may be described by this partial differ-ential equation:

∂2u

∂x2= ∂2u

∂t2− ε

∂4u

∂x2 ∂t2.

Show how to separate the variables.

8. Deflections of a thin plate and slow flow of a viscous fluid may both bedescribed by the biharmonic equation

∂4u

∂x4+ 2

∂4u

∂x2 ∂y2+ ∂4u

∂y4= 0.

Assume that u(x, y) = X(x)Y(y) and show that the variables don’t separate.Show that, under the additional assumption X′′/X = −λ2, a differentialequation for Y results.

4.7 Comments and ReferencesWhile the potential equation describes many physical phenomena, there is onethat makes the solution of the Dirichlet problem very easy to visualize. Sup-pose a piece of wire is bent into a closed curve or frame. When the frame isheld over a level surface, its projection onto the surface is a plane curve C en-closing a region R. If one forms a soap film on the frame, the height u(x, y) ofthe film above the level surface is a function that satisfies the potential equa-tion approximately, if the effects of gravity are negligible (see Chapter 5). Theheight of the frame above the curve C gives the boundary condition on u. Forexample, Fig. 2(b) shows the surface corresponding to the problem solved inSection 4.2. A great deal of information about soap films is in the book TheScience of Soap Films and Soap Bubbles by C. Isenberg (see the Bibliography).


Figure 3 Streamlines (solid) and equipotential curves (dashed) for flow in a cor-ner. The streamlines are described by the equation 2xy = constant, with a differ-ent constant for each one. Similarly, the equipotential curves are described by theequation x2 − y2 = constant.

It turns out that the potential equation (but not all elliptic equations) isbest studied through the use of complex variables. A complex variable may bewritten z = x + iy, where x and y are real and i2 = −1; similarly a function ofz is denoted by f (z) = u(x, y) + iv(x, y), u and v being real functions of realvariables. If f has a derivative with respect to z, then both u and v satisfy thepotential equation. Easy examples, such as polynomials and exponentials, leadto familiar solutions:

z2 = (x + iy)2 = x2 − y2 + i2xy,

ez = ex+iy = exeiy = ex cos(y) + iex sin(y),

ln(z) = 1

2ln

(x2 + y2

) + i tan−1

(y

x

).

(See Section 4.1, Exercises 1, 2; Section 4.4, Exercises 13, 14; MiscellaneousExercise 18 in this chapter.) Knowing these elementary solutions often helpsin simplifying a problem.

In certain idealized fluid flows (steady, irrotational, two-dimensional flow ofan inviscid, incompressible fluid) the velocity vector is given by V = − grad φ,where the velocity potential φ is a solution of the potential equation. Thestreamlines along which the fluid flows are level curves of a related function ψ ,called the stream function, which also is a solution of the potential equation.The two functions φ and ψ are, respectively, the real and imaginary partsof a function of the complex variable z. The level curves φ = constant andψ = constant form two families of orthogonal curves called a flow net. Theflow net in Fig. 3, for φ = x2 − y2 and ψ = 2xy (the real and imaginary partsof the function f (z) = z2), illustrates flow near a corner formed by two walls.Many other flow nets are shown in the book Potential Flows: Computer GraphicSolutions by R.H. Kirchhoff. Civil engineers sometimes sketch a flow net by eye


to get a rough graphical solution of the potential equation for hydrodynamicsproblems.

Where a physical boundary is formed by an impervious wall, the velocityvector V must be parallel to the boundary. This fact leads to two boundaryconditions. First, the wall must coincide with a streamline; thus ψ = constantalong a boundary. Second, the component of V that is normal to the wall mustbe zero there, because no fluid passes through it; thus the normal derivative ofφ is zero, ∂φ/∂n = 0, at a boundary. See Miscellaneous Exercises 30–32.

Chapter Review

See the CD for Review Questions.

Miscellaneous Exercises

1. Solve the potential equation in the rectangle 0 < x < a, 0 < y < b withthe boundary conditions

u(0, y) = 1, u(a, y) = 0, 0 < y < b,

u(x,0) = 0, u(x,b) = 0, 0 < x < a.

2. If a = b in Exercise 1, then u(a/2,a/2) = 1/4. Use symmetry to explainthis fact.

3. Solve the potential equation on the rectangle 0 < x < a, 0 < y < b withthe boundary conditions

u(0, y) = 1, u(a, y) = 1, 0 < y < b,

∂u

∂y(x,0) = 0,

∂u

∂y(x,b) = 0, 0 < x < a.

4. Same as Exercise 3, but the boundary conditions are

u(0, y) = 1,∂u

∂x(a, y) = 0, 0 < y < b,

u(x,0) = 1,∂u

∂y(x,b) = 0, 0 < x < a.



u(0, y) = 1, u(a, y) = 1, 0 < y < b,

∂u

∂y(x,0) = 0, u(x,b) = 0, 0 < x < a.


u(0, y) = 1, u(a, y) = 0, 0 < y < b,

u(x,0) = 1, u(x,b) = 0, 0 < x < a.

7. Same as Exercise 3, but the region is a square (b = a) and the boundaryconditions are

u(0, y) = f (y), u(a, y) = 0, 0 < y < a,

u(x,0) = f (x), u(x,a) = 0, 0 < x < a,

where f is a function whose graph is an isosceles triangle of height h andwidth a.

8. Solve the potential equation in the region 0 < x < a, 0 < y with theboundary conditions

u(x,0) = 1, 0 < x < a,

u(0, y) = 0, u(a, y) = 0, 0 < y.

9. Find the solution of the potential equation on the strip 0 < y < b,−∞ < x < ∞, subject to the conditions that follow. Supply bounded-ness conditions as necessary.

u(x,0) ={

1, −a < x < a,0, |x| > a,

u(x,b) = 0, −∞ < x < ∞.

10. Show that the function u(x, y) = tan−1(y/x) is a solution of the potentialequation in the first quadrant. What conditions does u satisfy along thepositive x- and y-axes?

11. Solve the potential problem in the upper half-plane,

∂2u

∂x2+ ∂2u

∂y2= 0, −∞ < x < ∞, 0 < y,

u(x,0) = f (x), −∞ < x < ∞,

taking f (x) = exp(−α|x|).


12. Apply the following formula (see Section 4.4, Exercise 16) for the solu-tion of the potential problem in the upper half-plane if the boundarycondition is u(x,0) = f (x), where

f (x) ={0, x < 0,

1, x > 0,

u(x, y) = 1

π

∫ ∞

−∞f (x′)

y

y2 + (x − x′)2dx′.

13. Apply the formula in Exercise 12 to the case where f (x) = 1, −∞ < x< ∞. The solution of the problem should be u(x, y) ≡ 1.

14. a. Find the separation-of-variables solution of the potential problem ina disk of radius 1 if the boundary condition is u(1, θ) = f (θ), where

f (θ) ={−π − θ, −π < θ < 0,

π − θ, 0 < θ < π .

b. Show that the function given in polar and Cartesian coordinates by

u(r, θ) = 2 tan−1

(r sin(θ)

1 − r cos(θ)

)

= 2 tan−1

(y

1 − x

)

satisfies the potential equation (use the Cartesian coordinates) andthe boundary condition. The following identity is useful:

sin(θ)

1 − cos(θ)= tan

(π − θ

2

).

c. Sketch some level curves of the solution inside the circle of radius 1.

15. Solve the potential equation in a disk of radius c with boundary condi-tions

u(c, θ) ={1, 0 < θ < π ,

0, −π < θ < 0.

16. What is the value of u at the center of the disk in Exercise 15?

17. Same as Exercise 15, but the boundary condition is

u(c, θ) = ∣∣sin(θ)∣∣.


18. For the potential problem on an annular ring

1

r

∂

∂r

(r∂2u

∂r

)+ 1

r2

∂2u

∂θ2= 0, a < r < b,

show that product solutions have the form

A0 + B0 ln(r), (C0 + D0θ) ln(r),

or

rn(An cos(nθ) + Bn sin(nθ)

) + r−n(Cn cos(nθ) + Dn sin(nθ)

).

19. Solve the potential problem on the annular ring as stated in Exercise 18with boundary conditions

u(a, θ) = 1, u(b, θ) = 0.

20. Find product solutions of the potential equation on a sector of a diskwith zero boundary conditions on the straight edges.

∇2u = 0, 0 ≤ r < c, 0 < θ < α,

u(r,0) = 0, u(r, α) = 0.

21. Solve the potential problem in a slit disk:

∇2u = 0, 0 ≤ r < c, 0 < θ < 2π,

u(r,0) = 0, u(r,2π) = 0,

u(r, θ) = f (θ), 0 < θ < 2π.

22. Show that the function u(x, y) = sin(πx/a) sinh(πy/a) satisfies the po-tential problem

∇2u = 0, 0 < x < a, 0 < y,

u(0, y) = 0, u(a, y) = 0, 0 < y,

u(x,0) = 0, 0 < x < a.

This solution is eliminated if it is also required that u(x, y) be boundedas y → ∞.

23. Solve the potential equation in the rectangle 0 < x < a, 0 < y < b, withthe boundary conditions

u(0, y) = 0,∂u

∂x(a, y) = 0, 0 < y < b,

u(x,0) = 0, u(x,b) = x, 0 < x < a.


24. Find a polynomial of second degree in x and y,

v(x, y) = A + Bx + Cy + Dx2 + Exy + Fy2,

that satisfies the potential equation and these boundary conditions:

v(0, y) = 0, 0 < y < b,

v(x,0) = 0, v(x,b) = x, 0 < x < a.

25. Find the problem (partial differential equation and boundary condi-tions) satisfied by w(x, y) = v(x, y) − u(x, y), where u and v are the so-lutions of the problems in Exercises 23 and 24. Solve the problem. Is thisproblem easier to solve than the one in Exercise 23?

26. Solve the potential equation in the quarter-plane 0 < x, 0 < y, subject tothe boundary conditions

u(x,0) = f (x), 0 < x,

u(0, y) = f (y), 0 < y.

The function f that appears in both boundary conditions is given by theequation

f (x) ={1, 0 < x < a,

0, a < x.

27. (Flow past a plate) A fluid occupies the half-plane y > 0 and flows past(left to right, approximately) a plate located near the x-axis. If the x and ycomponents of velocity are U0 + u(x, y) and v(x, y), respectively (U0 =constant free-stream velocity), under certain assumptions, the equationsof motion, continuity, and state can be reduced to

∂u

∂y= ∂v

∂x,

(1 − M2

)∂u

∂x+ ∂v

∂y= 0,

valid for all x and y > 0. M is the free-stream Mach number. Define thevelocity potential φ by the equations u = ∂φ/∂x and v = ∂φ/∂y. Showthat the first equation is automatically satisfied and the second is a partialdifferential equation that is elliptic if M < 1 or hyperbolic if M > 1.

28. If the plate is wavy — say, its equation is y = ε cos(αx) — then theboundary condition, that the vector velocity be parallel to the wall, is

v(x, ε cos(αx)

) = −εα sin(αx)(U0 + u

(x, ε cos(αx)

)).

This equation is impossible to use, so it is replaced by

v(x,0) = −εαU0 sin(αx)


on the assumption that ε is small and u is much smaller than U0. Usingthis boundary condition and the condition that u(x, y) → 0 as y → ∞,set up and solve a complete boundary value problem for φ, assumingM < 1.

29. By superposition of solutions (α ranging from 0 to ∞) find the flow pasta wall whose equation is y = f (x). Hint: Use the boundary condition

v(x,0) = U0f ′(x) =∫ ∞

0

[A(α) cos(αx) + B(α) sin(αx)

]dα.

30. In hydrodynamics, the velocity vector in a fluid is V = − grad(u), whereu is a solution of the potential equation. The normal component of ve-locity, ∂u/∂n, is 0 at a wall. Thus the problem

∇2u = 0, 0 < x < 1, 0 < y < 1,

∂u

∂x(0, y) = 0,

∂u

∂x(1, y) = −1, 0 < y < 1,

∂u

∂y(x,0) = 0,

∂u

∂y(x,1) = 1, 0 < x < 1,

represents a flow around a corner: flow inward at the top, outward at theright, with walls at left and bottom. Explain why, in a fluid flow problem,it must be true that ∫

C

∂u

∂nds = 0 (∗)

if u is a solution of the potential equation in a region R, ∂u/∂n is theoutward normal derivative, C is the boundary of the region, and s is arclength.

31. Under the conditions stated in Exercise 30, prove the validity of (∗).Hint: Use Green’s theorem.

32. The Neumann problem consists of the potential equation in a region Rand conditions on ∂u/∂n along C, the boundary of R. Show (a) that∫

C

∂u

∂nds = 0

is a necessary condition for a solution to exist, and (b) if u is a solutionof the Neumann problem, so is u + c (c is constant).

33. Show that u(x, y) = 12 (y2 −x2) is a solution of the problem in Exercise 30.

34. a. Show that the given function is a solution of the potential equation.


b. Find the gradient of u and plot some vectors V = − grad(u) near theorigin.

u(x, y) = − tan−1

(y

x

).

The flow field (see Exercise 30) given by this function is called anirrotational vortex.

35. Same tasks as in Exercise 34. The flow field given by this function is calleda source at the origin.

u(x, y) = − ln(√

x2 + y2).

36. Solve this potential problem in a half-annulus (sketch the region). Atsome point, it may be useful to make the substitution s = ln(r).

1

r

∂

∂r

(r∂u

∂r

)+ 1

r2

∂2u

∂θ2= 0, 1 < r < e, 0 < θ < π,

u(1, θ) = 0, u(e, θ) = 0, 0 < θ < π,

u(r,0) = 0, u(r,π) = 1, 1 < r < e.

37. Solve the Poisson equation, ∇2u = −f , in polar coordinates by finding afunction that depends only on r for:

a. f (r, θ) = 1;

b. f (r, θ) = 1

r2.

38. In “An improved transmission line structure for contact resistivity mea-surements” [L.P. Floyd et al., Solid-State Electronics, 37 (1994): 1579–1584], a strip of conducting material is carrying a current in the direc-tion of its length. A second, long conducting strip of width L is placed atright angles to the first, forming a cross. A voltage is to be measured bya probe on the second strip some distance from the first. In the secondstrip, the voltage V(x, y) satisfies the boundary value problem

∂2V

∂x2+ ∂2V

∂y2= 0, 0 < x < L, 0 < y,

∂V

∂x(0, y) = 0,

∂V

∂x(L, y) = 0, 0 < y,

V(x,0) = f (x), 0 < x < L.

In this problem, x is in the direction of current flow in the lower strip;y is in the direction of the length of the second strip; y = 0 at the edge


Figure 4 Exercise 38.

of the lower strip (see Fig. 4). Of course, V is bounded as y → ∞. Solvethis boundary value problem for V in terms of f (x).

39. The authors of the article cited in Exercise 38 say that any measurementof V(x, y) made at a distance y greater than 5L is independent of x. Ex-plain this statement, and determine what value (in terms of f ) would bemeasured.

40. In the article “A production-planning and design model for assessing thethermal behavior of thick steel strip during continuous heat treatment”[W.D. Morris, Journal of Process Engineering (2001): 53–63] the authormodels the temperature T(x, y) of a long steel strip that comes out ofan oven at x = 0, moving to the right, where it is exposed to coolant air.These equations enter into the modeling (see Table 1 and Fig. 5):

a. Conservation of energy/steady-state heat equation, derived by con-sidering conservation of energy for a rectangle of dimensions x byy that is fixed in space (see Sections 5.1 and 5.2):

v

k

∂T

∂x= ∂2T

∂x2+ ∂2T

∂y2, 0 < x, −b < y < b;

b. Symmetry condition:

∂T

∂y(x,0) = 0, 0 < x;

c. Cooling by convection at the surface (see Section 2.1, Eq. (10)):

−κ∂T

∂y(x,b) = h

(T(x,b) − Ta

), 0 < x;


h convection coefficient (W/m2K)κ thermal conductivity of steel (W/mK)L length of cooling lineT(x, y) temperature in the stripTa temperature of coolantT0 temperature of the strip at entry to cooling linev strip speed (m/s)k thermal diffusivity of steel (m2/s)B Biot number (dimensionless)

Table 1 Table of Notation.


d. Condition at entry to cooling line (at x = 0):

T(0, y) = T0, −b < y < b.

The author treats the strip as infinite. In fact, typical dimensions are100 m in the x-direction and 2 cm in the y-direction, so the ratio ofx to y lengths is on the order of 104.

Next, these dimensionless variables are introduced:

θ = T − Ta

T0 − Ta, Y = y

b, X = x

b,

and these dimensionless parameter combinations appear in the equa-tions:

γ = bv

k, B = hb

κ.


The problem in terms of dimensionless variables and parameters is:

γ∂θ

∂X= ∂2θ

∂X2+ ∂2θ

∂Y2, 0 < X, 0 < Y < 1,

∂θ

∂Y(X,0) = 0, 0 < X,

∂θ

∂Y(X,1) = −Bθ(X,1), 0 < X,

θ(0,Y) = 1, 0 < Y < 1.

Solve the problem for the case of a high Biot number, B → ∞, whichmeans that θ(X,1) = 0, 0 < X.

41. (Continuation) Solve the problem in Exercise 40 for the case of a lowBiot number, B ≈ 0, which means that ∂θ

∂Y (X,1) = 0, 0 < X.

Higher Dimensionsand OtherCoordinates

C H A P T E R

5

5.1 Two-Dimensional Wave Equation: Derivation

For an example of a two-dimensional wave equation, we consider a membranethat is stretched taut over a flat frame in the xy-plane (Fig. 1). The displace-ment of the membrane above the point (x, y) at time t is u(x, y, t). We assumethat the surface tension σ (dimensions F/L) is constant and independent ofposition. We also suppose that the membrane is perfectly flexible; that is, itdoes not resist bending. (A soap film satisfies these assumptions quite accu-rately.) Let us imagine that a small rectangle (of dimensions x by y alignedwith the coordinate axes) is cut out of the membrane, and then apply Newton’slaw of motion to it. On each edge of the rectangle, the rest of the membrane ex-erts a distributed force of magnitude σ (symbolized by the arrows in Fig. 2a);these distributed forces can be resolved into concentrated forces of magnitudeσ x or σ y, according to the length of the segment involved (see Fig. 2b andFig. 3).

Figure 1 Frame in the xy-plane.

295

296 Chapter 5 Higher Dimensions and Other Coordinates

(a) (b)

Figure 2 (a) Distributed forces. (b) Concentrated forces.

Figure 3 Forces on a piece of membrane.

(a) (b)

Figure 4 Forces (a) in the xu-plane; (b) in the yu-plane.

Looking at projection on the xu- and yu-planes (Figs. 4a, 4b), we see thatthe sum of forces in the x-direction is σ y(cos(β) − cos(α)), and the sum offorces in the y-direction is x(cos(δ) − cos(γ )). It is desirable that both thesesums be zero or at least negligible. Therefore we shall assume that α, β , γ , andδ are all small angles. Because we know that

tan(α) = ∂u

∂x, tan(γ ) = ∂u

∂y

and so forth, when the derivatives are evaluated at some appropriate point near(x, y), we are assuming that the slopes ∂u/∂x and ∂u/∂y of the membrane arevery small.

Chapter 5 Higher Dimensions and Other Coordinates 297

Adding up forces in the vertical direction and equating the sum to the masstimes acceleration (in the vertical direction) we obtain

σ y(sin(β) − sin(α)

) + σ x(sin(δ) − sin(γ )

) = ρ x y∂2u

∂t2,

where ρ is the surface density [m/L2]. Because the angles α,β, γ , and δ aresmall, the sine of each is approximately equal to its tangent:

sin(α) ∼= tan(α) = ∂u

∂x(x, y, t),

and so forth. With these approximations used throughout, the preceding equa-tion becomes

σy

(∂y

∂x(x + x, y, t) − ∂u

∂x(x, y, t)

)

+ σ x

(∂u

∂y(x, y + y, t) − ∂u

∂y(x, y, t)

)= ρ x y

∂2u

∂t2.

On dividing through by x y, we recognize two difference quotients inthe left-hand member. In the limit they become partial derivatives, yieldingthe equation

σ

(∂2u

∂x2+ ∂2u

∂y2

)= ρ

∂2u

∂t2,

or

∂2u

∂x2+ ∂2u

∂y2= 1

c2

∂2u

∂t2,

if c2 = σ/ρ. This is the two-dimensional wave equation.If the membrane is fixed to the flat frame, the boundary condition would be

u(x, y, t) = 0 for (x, y) on the boundary.

Naturally, it is necessary to give initial conditions describing the displacementand velocity of each point on the membrane at t = 0:

u(x, y,0) = f (x, y),

∂u

∂t(x, y,0) = g(x, y).

E X E R C I S E S

1. Suppose that the frame is rectangular, bounded by segments of the linesx = 0, x = a, y = 0, y = b. Write an initial value–boundary value problem,complete with inequalities, for a membrane stretched over this frame.


2. Suppose that the frame is circular and that its equation is x2 + y2 = a2.Write an initial value–boundary value problem for a membrane on a circu-lar frame. (Use polar coordinates.)

3. What should the three-dimensional wave equation be?

5.2 Three-Dimensional Heat Equation:Vector DerivationTo illustrate a different technique, we are going to derive the three-dimensionalheat equation using vector methods. Suppose we are investigating the temper-ature in a body that occupies a region R in space. (See Fig. 5.) Let V be asubregion of R bounded by the surface S . The law of conservation of energy,applied to V , says

net rate of heat in + rate of generation inside = rate of accumulation.

Our next job is to quantify this statement. The heat flow rate at any pointinside R is a vector function, q, measured in J/m2 s or similar units. The rate ofheat flow through a small piece of the surface S with area A is approximatelyn · q A (see Fig. 6), where n is the outward unit normal. This quantity ispositive for outward flow, so the inflow is its negative. The net inflow over theentire surface S is a sum of quantities like this, which becomes, in the limit asA shrinks, the integral ∫∫

S−q · n dA.

The term “rate of generation inside” in the energy balance is intended to in-clude conversion of energy from other forms (chemical, electrical, nuclear) tothermal. We assume that it is specified as an intensity g measured in J/m3 s or

Figure 5 A solid body occupying a region R in space and a subregion V withboundary S .

5.2 Three-Dimensional Heat Equation 299

Figure 6 The heat flow rate through a small section of surface with area A isq · n A.

similar units. Then the rate at which heat is generated in a small region of vol-ume V centered on point P is approximately g(P, t)V . These contributionsare summed over the whole subregion V ; as V shrinks, their total becomesthe integral ∫∫∫

Vg(P, t)dV .

The rate at which heat is stored in a small region of volume V centered onpoint P is proportional to the rate at which temperature changes there. Thatis, the storage rate is ρc V ut(P, t). The storage rate for the whole subregionV is the sum of such contributions, which passes to the integral∫∫∫

Vρc

∂u

∂t(P, t)dV .

Now the heat balance equation in mathematical terms becomes∫∫S

−q · n dA +∫∫∫

Vg dV =

∫∫∫V

ρc∂u

∂tdV . (1)

At this point, we call on the divergence theorem, which states that the integralover a surface S of the outward normal component of a vector function equalsthe integral over the volume bounded by S of the divergence of the function.Thus ∫∫

Sq · n dA =

∫∫∫V

∇ · q dV (2)

and we make this replacement in Eq. (1). Next collect all terms on one side ofthe equation to find

∫∫∫V

[−∇ · q + g − ρc

∂u

∂t

]dV = 0. (3)


Because the subregion V was arbitrary, we conclude that the integrand mustbe 0 at every point:

−∇ · q + g − ρc∂u

∂t= 0 in R, 0 < t. (4)

The argument goes this way. If the integrand were not identically 0, we couldfind some subregion of R throughout which it is positive (or negative). Theintegral over that subregion then would be positive (or negative), contradict-ing Eq. (3), which holds for any subregion.

The vector form of Fourier’s law of heat conduction says that the heat flowrate in an isotropic solid (same properties in all directions) is negatively pro-portional to the temperature gradient,

q = −κ ∇u. (5)

Again, the minus sign makes the heat flow “downhill” — from hotter to colderregions. Assuming that the conductivity κ is constant, we find, on substitutingFourier’s law into Eq. (4), the three-dimensional heat equation,

κ ∇2u + g = ρc∂u

∂tin R, 0 < t. (6)

Of course, we must add an initial condition of the form

u(P,0) = f (P) for P in R. (7)

In addition, at every point of the surface B bounding the region R, someboundary condition must be specified. Commonly we have conditions suchas those that follow, any one of which may be given on B or some portion ofit, B′.

(1) Temperature specified, u(P, t) = h1(P, t), for P any point in B′, whereh1 is a given function.

(2) Heat flow rate specified. The outward heat flow rate through a smallportion of surface surrounding point P on B′ is q(P, t) · n times the area. If thisis controlled, then by Fourier’s law ∇u · n is controlled. But this dot productis just the directional derivative of u in the outward normal direction at thepoint P. Thus, this type of boundary condition takes the form

∂u

∂n(P, t) = h2(P, t) for P on B′, (8)

where h2 is a given function.(3) Convection. If a part of the surface is exposed to a fluid at temperature

T(P, t), then an accounting of energy passing through a small piece of surfacecentered at P leads to the equation

q(P, t) · n = h(u(P, t) − T(P, t)

)for P on B′.

5.2 Three-Dimensional Heat Equation 301

Again using Fourier’s law, we obtain the boundary condition

κ∂u

∂n(P, t) + hu(P, t) = hT(P, t) for P on B′. (9)

As an example, we set up the three-dimensional problem for a solid in theform of a rectangular parallelepiped. In this case, Cartesian coordinates areappropriate, and we may describe the region R by the three inequalities 0 <

x < a, 0 < y < b, 0 < z < c. Assuming no generation inside the object, we havethe partial differential equation

∂2u

∂x2+ ∂2u

∂y2+ ∂2u

∂z2= 1

k

∂u

∂t, 0 < x < a, 0 < y < b, 0 < z < c, 0 < t.

(10)Suppose that on the faces at x = 0 and a, the temperature is controlled, so

the boundary condition there is

u(0, y, z, t) = T0, u(a, y, z, t) = T1, 0 < y < b, 0 < z < c, 0 < t.(11)

Furthermore, assume that the top and bottom surfaces are insulated. Then theboundary conditions at z = 0 and c are

∂u

∂z(x, y,0, t) = 0,

∂u

∂z(x, y, c, t) = 0, 0 < x < a, 0 < y < b, 0 < t.

(12)(The outward normal directions on the top and bottom are the positive andnegative z-directions, respectively.) Finally, assume that the faces at y = 0 andat y = b are exposed to a fluid at temperature T2, so they transfer heat byconvection there. The resulting boundary conditions are

−κ∂u

∂y(x,0, z, t) + hu(x,0, z, t) = hT2,

κ∂u

∂y(x,b, z, t) + hu(x,b, z, t) = hT2,

0 < x < a, 0 < z < c, 0 < t. (13)

Finally, we add an initial condition,

u(x, y, z,0) = f (x, y, z), 0 < x < a, 0 < y < b, 0 < z < c. (14)

A full, three-dimensional problem is complicated to solve, so we often lookfor ways to reduce it to two or even one dimension. In the example problemof Eqs. (10)–(14), we might eliminate z by finding the temperature averagedover the interval 0 < z < c,

v(x, y, t) = 1

c

∫ c

0u(x, y, z, t)dz.


Because differentiation with respect to x, y, or t gives the same result insideor outside the integral with respect to z, and because of the boundary condi-tion (12), we find that

1

c

∫ c

0

(∂2u

∂x2+ ∂2u

∂y2+ ∂2u

∂z2

)dz = ∂2v

∂x2+ ∂2v

∂y2,

and v satisfies the two-dimensional heat equation,

∂2v

∂x2+ ∂2v

∂y2= 1

k

∂v

∂t, 0 < x < a, 0 < y < b, 0 < t.

(See the exercises for details and for boundary and initial conditions.)If z-variation cannot be ignored, we could try to get rid of the y-variation

by introducing an average in that direction,

w(x, z, t) = 1

b

∫ b

0u(x, y, z, t)dy.

From the boundary condition (13) we find that

∫ b

0

∂2u

∂y2dy = ∂u

∂y(x,b, z, t) − ∂u

∂y(x,0, z, t)

=(

h

κ

)[(T2 − u(x,b, z, t)

) + (T2 − u(x,0, z, t)

)].

If b is small — the parallelepiped is more like a plate — we may accept the ap-proximation u(x,b, z, t) + u(x,0, z, t) ≡ 2w(x, z, t), which would make, fromthe preceding expression,

1

b

∫ b

0

∂2u

∂y2dy ≡

(2h

bκ

)(T2 − w(x, z, t)

).

After applying the averaging process to Eqs. (10), (11), (12), and (14) we ob-tain the following two-dimensional problem for w:

∂2w

∂x2+ ∂2w

∂z2+ 2h

bκ(T2 − w) = 1

k

∂w

∂t, 0 < x < a, 0 < z < c, 0 < t,

(15)

w(0, z, t) = T0, w(a, z, t) = T1, 0 < z < c, 0 < t, (16)

∂w

∂z(x,0, t) = 0,

∂w

∂z(x, c, t) = 0, 0 < x < a, 0 < t, (17)

w(x, z,0) = 1

b

∫ b

0f (x, y, z)dy, 0 < x < a, 0 < z < c. (18)

5.3 Two-Dimensional Heat Equation: Solution 303

E X E R C I S E S

1. For the function u(x, y, z, t) that satisfies Eqs. (10)–(14), show that∫ c

0

∂2u

∂z2dz = 0.

2. Find the initial and boundary conditions satisfied by the function

v(x, y, t) = 1

c

∫ c

0u(x, y, z, t)dz,

where u satisfies Eqs. (10)–(14).

3. In Eqs. (15)–(18), suppose that w(x, z, t) → W(x, z) as t → ∞. State andsolve the boundary value problem for W . (This problem is much easierthan it appears, because there is no variation with z.)

4. Find the dimensions of ρ, c, κ,q, and g, and verify that the dimensions ofthe right and left members of the heat equation are the same.

5. Suppose the plate lies in the rectangle 0 < x < a, 0 < y < b. State a completeinitial value–boundary value problem for temperature in the plate if: thereis no heat generation; the temperature is held at T0 along x = a and y = 0;the edges at x = 0 and y = b are insulated.

5.3 Two-Dimensional Heat Equation:Double Series SolutionIn order to see the technique of solution for a two-dimensional problem, weshall consider the diffusion of heat in a rectangular plate of uniform, isotropicmaterial. The steady-state temperature distribution is a solution of the poten-tial equation (see Exercise 6). Suppose that the initial value–boundary valueproblem for the transient temperature u(x, y, t) is

∂2u

∂x2+ ∂2u

∂y2= 1

k

∂u

∂t, 0 < x < a, 0 < y < b, 0 < t, (1)

u(x,0, t) = 0, u(x,b, t) = 0, 0 < x < a, 0 < t, (2)

u(0, y, t) = 0, u(a, y, t) = 0, 0 < y < b, 0 < t, (3)

u(x, y,0) = f (x, y), 0 < x < a, 0 < y < b. (4)

This problem contains a homogeneous partial differential equation and ho-mogeneous boundary conditions. We may thus proceed with separation of


variables by seeking solutions in the form

u(x, y, t) = φ(x, y)T(t).

On substituting u in product form into Eq. (1), we find that it becomes(∂2φ

∂x2+ ∂2φ

∂y2

)T = 1

kφT ′.

Separation can be achieved by dividing through by φT, which leaves(∂2φ

∂x2+ ∂2φ

∂y2

)1

φ= T ′

kT.

We may argue, as usual, that the common value of the members of this equa-tion must be a constant, which we expect to be negative (−λ2). The equationsthat result are

T ′ + λ2kT = 0, 0 < t, (5)

∂2φ

∂x2+ ∂2φ

∂y2= −λ2φ, 0 < x < a, 0 < y < b. (6)

In terms of the product solutions, the boundary conditions become

φ(x,0)T(t) = 0, φ(x,b)T(t) = 0,

φ(0, y)T(t) = 0, φ(a, y)T(t) = 0.

In order to satisfy all four equations, either T(t) ≡ 0 for all t or φ = 0 on theboundary. We have seen many times that the choice of T(t) ≡ 0 wipes out oursolution completely. Therefore, we require that φ satisfy the conditions

φ(x,0) = 0, φ(x,b) = 0, 0 < x < a, (7)

φ(0, y) = 0, φ(a, y) = 0, 0 < y < b. (8)

We are not yet out of difficulty, because Eqs. (6)–(8) constitute a new prob-lem, a two-dimensional eigenvalue problem. It is evident, however, that thepartial differential equation and the boundary conditions are linear and ho-mogeneous; thus separation of variables may work again. Supposing that φ

has the form

φ(x, y) = X(x)Y(y),

we find that the partial differential equation (6) becomes

X′′(x)

X(x)+ Y ′′(y)

Y(y)= −λ2, 0 < x < a, 0 < y < b.


The sum of a function of x and a function of y can be constant only if thosetwo functions are individually constant:

X′′

X= constant,

Y ′′

Y= constant.

Before naming the constants, let us look at the boundary conditions onφ = XY :

X(x)Y(0) = 0, X(x)Y(b) = 0, 0 < x < a,

X(0)Y(y) = 0, X(a)Y(y) = 0, 0 < y < b.

If either of the functions X or Y is zero throughout the whole interval of itsvariable, the conditions are certainly satisfied, but φ is identically zero. Wetherefore require each of the functions X and Y to be zero at the endpoints ofits interval:

Y(0) = 0, Y(b) = 0, (9)

X(0) = 0, X(a) = 0. (10)

Now it is clear that each of the ratios X′′/X and Y ′′/Y should be a negativeconstant, designated by −µ2 and −ν2, respectively. The separate equations forX and Y are

X′′ + µ2X = 0, 0 < x < a, (11)

Y ′′ + ν2Y = 0, 0 < y < b. (12)

Finally, the original separation constant −λ2 is determined by

λ2 = µ2 + v2. (13)

Now we see two independent eigenvalue problems: Eqs. (9) and (12) formone problem and Eqs. (10) and (11) the other. Each is of a very familiar form;the solutions are

Xm(x) = sin

(mπx

a

), µ2

m =(

mπ

a

)2

, m = 1,2, . . . ,

Yn(y) = sin

(nπy

b

), ν2

n =(

nπ

b

)2

, n = 1,2, . . . .

Notice that the indices n and m are independent. This means that φ will havea double index. Specifically, the solutions of the two-dimensional eigenvalueproblem Eqs. (6)–(8) are

φmn(x, y) = Xm(x)Yn(y),

λ2mn = µ2

m + ν2n,


and the corresponding function T is

Tmn = exp(−λ2

mnkt).

We now begin to assemble the solution. For each pair of indices m, n (m =1,2,3, . . . , n = 1,2,3, . . .) there is a function

umn(x, y, t) = φmn(x, y)Tmn(t)

= sin

(mπx

a

)sin

(nπy

b

)exp

(−λ2mnkt

)that satisfies the partial differential equation (1) and the boundary conditionsEqs. (2) and (3). We may form linear combinations of these solutions to getother solutions. The most general linear combination would be the doubleseries

u(x, y, t) =∞∑

m=1

∞∑n=1

amnφmn(x, y)Tmn(t), (14)

and any such combination should satisfy Eqs. (1)–(3). There remains the ini-tial condition Eq. (4) to be satisfied. If u has the form given in Eq. (14), thenthe initial condition becomes

∞∑m=1

∞∑n=1

amnφmn(x, y) = f (x, y), 0 < x < a, 0 < y < b. (15)

The idea of orthogonality is once again applicable to the problem of selectingthe coefficients amn. One can show by direct computation that

∫ b

0

∫ a

0φmn(x, y)φpq(x, y)dx dy =

ab

4if m = p and n = q,

0, otherwise.(16)

Thus, the appropriate formula for the coefficients amn is

amn = 4

ab

∫ b

0

∫ a

0f (x, y) sin

(mπx

a

)sin

(nπy

b

)dx dy. (17)

If f is a sufficiently regular function, the series in Eq. (15) will converge andequal f (x, y) in the rectangular region 0 < x < a, 0 < y < b. We may then saythat the problem is solved. It is reassuring to notice that each term in the seriesof Eq. (14) contains a decaying exponential, and thus, as t increases, u(x, y, t)tends to zero, as expected.

Example.Let us take the specific initial condition

f (x, y) = xy, 0 < x < a, 0 < y < b.


The coefficients are easily found to be

amn = 4ab

π2

cos(mπ) cos(nπ)

mn= 4ab

π2

(−1)m+n

mn,

so the solution to this problem is

u(x, y, t) = 4ab

π2

∞∑m=1

∞∑n=1

(−1)m+n

mnsin

(mπx

a

)sin

(nπy

b

)exp

(−λ2mnkt

).

(18)This solution is shown animated on the CD. �

The double series that appear here are best handled by converting them intosingle series. To do this, arrange the terms in order of increasing values of λ2

mn.Then the first terms in the single series are the most significant, those thatdecay least rapidly. For example, if a = 2b, so that

λ2mn = (m2 + 4n2)π2

a2,

then the following list gives the double index (m,n) in order of increasingvalues of λ2

mn:

(1,1), (2,1), (3,1), (1,2), (2,2), (4,1), (3,2), . . . .

E X E R C I S E S

1. Write out the “first few” terms of the series of Eq. (18). By “first few,”we mean those for which λ2

mn is smallest. (Assume a = b in determiningrelative magnitudes of the λ2.)

2. Provide the details of the separation of variables by which Eqs. (9)–(13)are derived.

3. Find the frequencies of vibration of a rectangular membrane. See Sec-tion 5.1, Exercise 1.

4. Verify that umn(x, y, t) satisfies Eqs. (1)–(3).

5. Show that Xm(x) = cos(mπx/a) (m = 0,1,2, . . .) if the boundary condi-tions Eq. (3) are replaced by

∂u

∂x(0, y, t) = 0,

∂u

∂x(a, y, t) = 0, 0 < y < b, 0 < t.

What values will the λ2mn have, and of what form will the solution u(x, y, t)

be?


6. Suppose that, instead of boundary conditions Eqs. (2) and (3), we have

u(x,0, t) = f1(x), u(x,b, t) = f2(x), 0 < x < a, 0 < t, (2′)u(0, y, t) = g1(y), u(a, y, t) = g2(y), 0 < y < b, 0 < t. (3′)

Show that the steady-state solution involves the potential equation, andindicate how to solve it.

7. Solve the two-dimensional heat conduction problem in a rectangle if thereis insulation on all boundaries and the initial condition is

a. u(x, y,0) = 1;

b. u(x, y,0) = x + y;

c. u(x, y,0) = xy.

8. Verify the orthogonality relation in Eq. (16) and the formula for amn.

9. Show that the separation constant −λ2 must be negative by showing that−µ2 and −ν2 must both be negative.

10. Show that the function

umn(x, y, t) = sin(µmx) sin(νny) cos(λmnct),

where µm, νn, and λmn are as in this section, is a solution of the two-dimensional wave equation on the rectangle 0 < x < a, 0 < y < b, withu = 0 on the boundary. The function u may be thought of as the displace-ment of a rectangular membrane (see Section 5.1).

11. The places where umn(x, y, t) = 0 for all t are called nodal lines. Describethe nodal lines for

(m,n) = (1,2), (2,3), (3,2), (3,3).

12. Determine the frequencies of vibration for the functions umn of Exer-cise 10. Are there different pairs (m,n) that have the same frequency ifa = b?

5.4 Problems in Polar CoordinatesWe found that the one-dimensional wave and heat problems have a great dealin common. Namely, the steady-state or time-independent solutions and theeigenvalue problems that arise are identical in both cases. Also, in solvingproblems in a rectangular region, we have seen that those same features areshared by the heat and wave equations.

5.4 Problems in Polar Coordinates 309

If we consider now the vibrations of a circular membrane or heat conduc-tion in a circular plate, we shall see common features again. In what follows,these two problems are given side by side for the region 0 < r < a, 0 < t.

Wave Heat

∇2v = 1

c2

∂2v

∂t2∇2v = 1

k

∂v

∂t

v(a, θ, t) = f (θ) v(a, θ, t) = f (θ)

v(r, θ,0) = g(r, θ) v(r, θ,0) = g(r, θ)∂v

∂t(r, θ,0) = h(r, θ)

In both problems we require that v be periodic in θ with period 2π :

v(r, θ, t) = v(r, θ + 2π, t),

as in Section 4.5.Although the interpretation of the function v is different in the two cases,

we see that the solution of the problem

∇2v = 0, v(a, θ) = f (θ)

is the rest-state or steady-state solution for both problems, and it will beneeded in both problems to make the boundary condition at r = a homo-geneous. Let us suppose that the time-independent solution has been foundand subtracted; that is, we will replace f (θ) by zero. Then we have

∇2v = 1

c2

∂2v

∂t2∇2v = 1

k

∂v

∂t

v(a, θ, t) = 0 v(a, θ, t) = 0

plus the appropriate initial conditions. If we attempt to solve by separation ofvariables, setting v(r, θ, t) = φ(r, θ)T(t), in both cases we will find that φ(r, θ)

must satisfy

1

r

∂

∂r

(r∂φ

∂r

)+ 1

r2

∂2φ

∂θ2= −λ2φ, (1)

φ(a, θ) = 0, (2)

φ(r, θ + 2π) = φ(r,π), (3)

φ bounded as r → 0. (4)

Now we shall concentrate on the solution of this two-dimensional eigen-value problem. We can separate variables again by assuming that φ(r, θ) =


R(r)Q(θ). After some algebra, we find that

(rR′)′

rR+ Q′′

r2Q= −λ2, (5)

R(a) = 0, (6)

Q(θ) = Q(θ + 2π), (7)

R(r) bounded as r → 0. (8)

The ratio Q′′/Q must be constant; otherwise, λ2 could not be constant. Choos-ing Q′′/Q = −µ2, we get a familiar, singular eigenvalue problem:

Q′′ + µ2Q = 0, (9)

Q(θ + 2π) = Q(θ). (10)

We found (in Chapter 4) that the solutions of this problem are

µ20 = 0, Q0(θ) = 1,

µ2m = m2, Qm(θ) = cos(mθ) and sin(mθ),

(11)

where m = 1,2,3, . . . .There remains a problem in R:

(rR′)′ − µ2

rR + λ2rR = 0, 0 < r < a, (12)

R(a) = 0, (13)

R(r) bounded as r → 0. (14)

Equation (12) is called Bessel’s equation, and we shall solve it in the next sec-tion.

E X E R C I S E S

1. State the full initial value–boundary value problems that result from theproblems as originally given when the steady-state or time-independent so-lution is subtracted from v.

2. Verify the separation of variables that leads to Eqs. (1) and (2).

3. Substitution of v(r, θ, t) in the form of a product led to the problem ofEqs. (1)–(4) for the factor φ(r, θ). What differential equation is to be satis-fied by the factor T(t)?

4. Solve Eqs. (9)–(11) and check the solutions given.

5.5 Bessel’s Equation 311

5. Suppose the problems originally stated were to be solved in the half-disk0 < r < a, 0 < θ < π , with additional conditions:

v(r,0, t) = 0, 0 < r < a, 0 < t,

v(r,π, t) = 0, 0 < r < a, 0 < t.

What eigenvalue problem arises in place of Eqs. (9)–(11)? Solve it.

6. Suppose that the boundary condition

∂v

∂r(a, θ, t) = 0, −π < θ ≤ π, 0 < t

were given instead of v(a, θ, t) = f (θ). Carry out the steps involved in sepa-ration of variables. Show that the only change is in Eqs. (6) and (13), whichbecome R′(a) = 0.

7. One of the consequences of Green’s theorem is the integral relation∫∫R

(f ∇2g − g∇2f

)dA =

∫C

(f∂g

∂n− g

∂f

∂n

)ds,

where R is a region in the plane, C is the closed curve that bounds R, and∂f /∂n is the directional derivative in the direction normal to the curve C.Use this relation to show that eigenfunctions of the problem

∇2φ = −λ2φ in R,

φ = 0 on C

are orthogonal if they correspond to different eigenvalues. (Hint: Usef = φk, g = φm, m �= k.)

8. Same problem as Exercise 7, except the boundary condition is

φ + λ∂φ

∂n= 0 on C.

5.5 Bessel’s EquationIn order to solve the Bessel equation,

(rR′)′ − µ2

rR + λ2rR = 0, (1)

we apply the method of Frobenius. Assume that R(r) has the form of a powerseries multiplied by an unknown power of r:

R(r) = rα(c0 + c1r + · · · + ckrk + · · · ). (2)


When the differentiations in Eq. (1) are carried out and the equation is multi-plied by r, it becomes

r2R′′ = α(α − 1)c0rα + (α + 1)αc1rα+1 + (α + 2)(α + 1)c2rα+2 + · · ·+ (α + k)(α + k − 1)ckrα+k + · · ·

rR′ = αc0rα + (α + 1)c1rα+1 + (α + 2)c2rα+2 + · · ·+ (α + k)ckrα+k + · · ·

− µ2R = − µ2c0rα − µ2c1rα+1 − µ2c2rα+2 − · · ·− µ2ckrα+k + · · ·

λ2r2R = λ2c0rα+2 + · · ·+ λ2ck−2rα+k + · · ·

The expression for λ2r2R is jogged to the right to make like powers of r line upvertically. Note that the lowest power of r present in λ2r2R is rα+2.

Now we add the tableau vertically. The sum of the left-hand sides is, accord-ing to the differential equation, equal to zero. Therefore

0 = c0

(α2 − µ2

)rα + c1

[(α + 1)2 − µ2

]rα+1

+ [c2

((α + 2)2 − µ2

) + λ2c0

]rα+2

+ · · · + [ck

((α + k)2 − µ2

) + λ2ck−2

]rα+k + · · · .

Each term in this power series must be zero in order for the equality to hold.Therefore, the coefficient of each term must be zero:

c0

(α2 − µ2

) = 0,

c1

((α + 1)2 − µ2

) = 0,

...ck

((α + k)2 − µ2

) + λ2ck−2 = 0, k ≥ 2.

As a bookkeeping agreement, we take c0 �= 0. Thus α = ±µ. Let us study thecase α = µ ≥ 0. The second equation becomes

c1

((µ + 1)2 − µ2

) = 0

and this implies c1 = 0. Now, in general the relation

ck = − λ2ck−2

(µ + k)2 − µ2= −λ2 ck−2

k(2µ + k), k ≥ 2, (3)

says that ck can be found from ck−2. In particular, we find

c2 = − λ2

2(2µ + 2)c0,

c4 = − λ2

4(2µ + 4)c2 = λ4

2 · 4 · (2µ + 2)(2µ + 4)c0,


and so forth. All c’s with odd index are zero, since they are all multiples of c1.The general formula for a coefficient with even index k = 2m is

c2m = (−1)m

m!(µ + 1)(µ + 2) · · · (µ + m)

(λ

2

)2m

c0. (4)

For integral values of µ, c0 is chosen by convention to be

c0 =(

λ

2

)µ

· 1

µ!.

Then the solution of Eq. (1) that we have found is called the Bessel function ofthe first kind of order µ:

Jµ(λr) =(

λr

2

)µ ∞∑m=0

(−1)m

m!(µ + m)!

(λr

2

)2m

. (5)

This series serves us for evaluating the function and for obtaining its proper-ties. (See the Exercises.) However, from now on, we consider the Bessel functionsof the first kind to be as well known as sines and cosines, although less familiar.

There must be a second independent solution of Bessel’s equation, whichcan be found by using variation of parameters. This method yields a solutionin the form

Jµ(λr) ·∫

dr

r J2µ(λr)

. (6)

In its standard form, the second solution of Bessel’s equation is called the Besselfunction of second kind of order µ and is denoted by Yµ(λr).

The most important feature of the second solution is its behavior near r = 0.When r is very small, we can approximate Jµ(λr) by the first term of its seriesexpansion:

Jµ(λr) ∼=(

λ

2

)µ 1

µ!rµ, r 1.

The solution Eq. (6) then can be approximated by

constant × rµ

∫dr

r1+2µ= constant ×

{ln(r), if µ = 0,r−µ, if µ > 0.

In either case, it is easy to see that∣∣Yµ(λr)∣∣ → ∞ as r → 0.

Both kinds of Bessel functions have an infinite number of zeros. That is,there is an infinite number of values of α (and β) for which

Jµ(α) = 0, Yµ(β) = 0.


Figure 7 Graphs of Bessel functions of the first kind. Also see the CD. (a) J0 andJ1, (b) Y0 and Y1.

nm 1 2 3 40 2.405 5.520 8.654 11.7921 3.832 7.016 10.173 13.3242 5.136 8.417 11.620 14.7963 6.380 9.761 13.015 16.223

Table 1 Zeros of Bessel functions. The values αmn

satisfy the equation Jm(αmn) = 0.

Also, as r → ∞, both Jµ(λr) and Yµ(λr) tend to zero. Figure 7 gives graphsof several Bessel functions, and Table 1 provides values of their zeros. Furtherinformation can be found in most books of tables.

The modified Bessel equation differs from the Bessel equation only in thesign of one term. It is

(rR′)′ − µ2

rR − λ2rR = 0. (7)


Using the same method as in the preceding, an infinite series can be developedfor the solutions (see Exercise 8). The solution that is bounded at r = 0, instandard form, is called the modified Bessel function of the first kind of order µ,designated Iµ(λr), and its series is

Iµ(λr) =(

λr

2

)µ ∞∑m=0

1

m!(µ + m)!

(λr

2

)2m

.

Summary

The differential equation

d

dr

(r

dR

dr

)− µ2

rR + λ2r R = 0

is called Bessel’s equation. Its general solution is

R(r) = AJµ(λr) + BYµ(λr)

(A and B are arbitrary constants). The functions Jµ and Yµ are calledBessel functions of order µ of the first and second kinds, respectively.The Bessel function of the second kind is unbounded at the origin.

E X E R C I S E S

1. Find the values of the parameter λ for which the following problem has anonzero solution:

1

r

d

dr

(r

dφ

dr

)+ λ2φ = 0, 0 < r < a,

φ(a) = 0, φ(0) bounded.

2. Sketch the first few eigenfunctions found in Exercise 1.

3. Show that

d

drJµ(λr) = λJ ′

µ(λr),

where the prime denotes differentiation with respect to the argument.

4. Show from the series that

d

drJ0(λr) = −λJ1(λr).


5. By using Exercise 4 and Rolle’s theorem, and knowing that J0(x) = 0 for aninfinite number of values of x, show that J1(x) = 0 has an infinite numberof solutions.

6. Using the infinite series representations for the Bessel functions, verify theformulas

d

dx

(x−µJµ(x)

) = −x−µJµ+1(x),

d

dx

(xµJµ(x)

) = xµJµ−1(x).

7. Use the second formula in Exercise 6 to derive the integral formula∫xµJµ(x)x dx = xµ+1Jµ+1(x).

8. Use the method of Frobenius to obtain a solution of the modified Besselequation (7). Show that the coefficients of the power series are just thesame as those for the Bessel function of the first kind, except for signs.

9. Use the modified Bessel function to solve this problem for the temperaturein a circular plate when the surface is exposed to convection:

1

r

d

dr

(r

du

dr

)− γ 2(u − T) = 0, 0 < r < a,

u(a) = T1.

10. Using the result of Exercise 4, solve the eigenvalue problem

1

r

d

dr

(r

dφ

dr

)+ λ2φ = 0, 0 < r < a,

dφ

dr(a) = 0, φ(0) bounded.

5.6 Temperature in a CylinderIn Section 5.4, we observed that both the heat and wave equations have a greatdeal in common, especially the equilibrium solution and the eigenvalue prob-lem. To reinforce that observation, we will solve a heat problem and a waveproblem with analogous conditions so that their similarities may be seen.These examples illustrate another important point: Problems that would betwo-dimensional in one coordinate system (rectangular) may become one-dimensional in another system (polar). In order to obtain this simplification,we will assume that the unknown function, v(r, θ, t), is actually independent

5.6 Temperature in a Cylinder 317

of the angular coordinate θ . (We write v(r, t) then.) As a consequence of thisassumption, the two-dimensional Laplacian operator becomes

∇2v = 1

r

∂

∂r

(r∂v

∂r

).

Suppose that the temperature v(r, t) in a large cylinder (radius a) satisfiesthe problem

1

r

∂

∂r

(r∂v

∂r

)= 1

k

∂v

∂t, 0 < r < a, 0 < t, (1)

v(a, t) = 0, 0 < t, (2)

v(r,0) = f (r), 0 < r < a. (3)

Because the differential equation (1) and boundary condition (2) are ho-mogeneous, we may start the separation of variables by assuming v(r, t)=φ(r)T(t). Using this form for v, we find that the partial differential equa-tion (1) becomes

1

r(rφ′)′T = 1

kφT ′.

After dividing through this equation by φT, we arrive at the equality

(rφ′(r))′

rφ(r)= T ′(t)

kT(t). (4)

The two members of this equation must both be constant; call their mutualvalue −λ2. Then we have two linked ordinary differential equations,

T ′ + λ2kT = 0, 0 < t, (5)

(rφ′)′ + λ2rφ = 0, 0 < r < a. (6)

The boundary condition, Eq. (2), becomes φ(a)T(t) = 0, 0 < t. It will be sat-isfied by requiring that

φ(a) = 0. (7)

We can recognize Eq. (6) as Bessel’s equation with µ = 0. (See Summary,Section 5.5.) The general solution, therefore, has the form

φ(r) = AJ0(λr) + BY0(λr).

If B �= 0, φ(r) must become infinite as r approaches zero. The physical impli-cations of this possibility are unacceptable, so we require that B = 0. In effectwe have added the boundedness condition∣∣v(r, t)

∣∣ bounded at r = 0, (8)

which we shall employ frequently.


The function φ(r) = J0(λr) is a solution of Eq. (6), and we wish to choose λ

so that Eq. (7) is satisfied. Then we must have

J0(λa) = 0

or

λn = αn

a, n = 1,2, . . . ,

where αn are the zeros of the function J0. Thus the eigenfunctions and eigen-values of Eqs. (6), (7), and (8) are

φn(r) = J0(λnr), λ2n =

(αn

a

)2

. (9)

These are shown on the CD.Returning to Eq. (5), we determine that the time factors Tn are

Tn(t) = exp(−λ2

nkt).

We may now assemble the general solution of the partial differential equa-tion (1), under the boundary condition (2) and boundedness condition (8),as a general linear combination of our product solutions:

v(r, t) =∞∑

n=1

an J0(λnr) exp(−λ2

nkt). (10)

It remains to determine the coefficients an so as to satisfy the initial condi-tion (3), which now takes the form

v(r,0) =∞∑

n=1

an J0(λnr) = f (r), 0 < r < a. (11)

While this problem is not a routine exercise in Fourier series or even a reg-ular Sturm–Liouville problem (see Section 2.7, especially Exercise 6 there), itis nevertheless true that the eigenfunctions of Eqs. (6) and (7) are orthogonal,as expressed by the relation∫ a

0φn(r)φm(r)r dr = 0 (n �= m)

or ∫ a

0J0(λnr)J0(λmr)r dr = 0 (n �= m).

More importantly, the following theorem gives us justification for Eq. (11).

5.6 Temperature in a Cylinder 319

Theorem. If f (r) is sectionally smooth on the interval 0 < r < a, then at everypoint r on that interval,

∞∑n=1

an J0(λnr) = f (r+) + f (r−)

2, 0 < r < a,

where the λn are solutions of J0(λa) = 0 and

an =∫ a

0 f (r)J0(λnr)r dr∫ a0 J2

0(λnr)r dr. (12)

�

The CD shows an animation of a Bessel series converging.Now we may proceed with the problem at hand. If the function f (r) in the

initial condition (3) is sectionally smooth, the use of Eq. (12) to chose thecoefficients an guarantees that Eq. (11) is satisfied (as nearly as possible), andhence the function

v(r, t) =∞∑

n=1

an J0(λnr) exp(−λ2

nkt)

(13)

satisfies the problem expressed by Eqs. (1), (2), (3), and (8).By way of example, let us suppose that the function f (r) = T0, 0 < r < a. It

is necessary to determine the coefficients an by formula (12). The numeratoris the integral ∫ a

0T0 J0(λnr)r dr.

This integral is evaluated by means of the relation (see Exercise 6 of Sec-tion 5.5)

d

dx

(x J1(x)

) = x J0(x). (14)

Hence, we find ∫ a

0J0(λnr)r dr = 1

λnr J1(λnr)

∣∣∣∣a

0

= a

λnJ1(λna) = a2

αnJ1(αn). (15)

The denominator of Eq. (12) is known to have the value (Exercise 5)∫ a

0J2

0(λnr)r dr = a2

2J2

1(λna)

= a2

2J2

1(αn). (16)


n αn J1(αn)2

αn J1(αn)

1 2.405 +0.5191 +1.60202 5.520 −0.3403 −1.06473 8.654 +0.2715 +0.85124 11.792 −0.2325 −0.7295

Table 2 Values for Eq. (18)

Figure 8 Graphs of the solution of the example problem. The function v(r, t) asgiven in Eq. (18) is shown vs r for times chosen so that kt/a2 takes the values 0,0.01, 0.1, and 0.5. T0 = 100 and a = 1.

Putting together the numerator and denominator from Eqs. (15) and (16),we find that the coefficients we need are

an = 2T0

αn J1(αn). (17)

Thus, the solution to the heat conduction problem is

v(r, t) = T0

∞∑n=1

2

αn J1(αn)J0(λnr) exp

(−λ2nkt

). (18)

In Table 2 are listed the first few values of the ratio 2/[αn J1(αn)]. Figure 8shows graphs of v(r, t) as a function of r for several times. Also, see Exercise 1.An animation is shown on the CD.

5.7 Vibrations of a Circular Membrane 321

E X E R C I S E S

1. Use Eq. (18) to find an expression for the function v(0, t)/T0. Evaluate thefunction for

kt

a2= 0.1,0.2,0.3.

(The first two terms of the series are sufficient.)

2. Write out the first three terms of the series in Eq. (18).

3. Solve the heat problem consisting of Eqs. (1)–(3) if f (r) is

f (r) =

T0, 0 < r <a

2,

0,a

2< r < a.

4. Let φ(r) = J0(λr) so that φ(r) satisfies Bessel’s equation of order 0. Multiplythrough the differential equation by rφ′ and conclude that

d

dr

[(rφ′)2

] + λ2r2 d

dr

[φ2

] = 0.

5. Assuming that λ is chosen so that φ(a) = 0, integrate the equation in Exer-cise 4 over the interval 0 < r < a to find∫ a

0φ2(r)r dr = 1

2λ2

(aφ′(a)

)2.

6. Use Exercise 5 to validate Eq. (16).

5.7 Vibrations of a Circular MembraneWe shall now attempt to solve the problem of describing the displacement of acircular membrane that is fixed at its edge.

Symmetric VibrationsTo begin with, we treat the simple case in which the initial conditions are in-dependent of θ . Thus the displacement v(r, t) satisfies the problem

1

r

∂

∂r

(r∂v

∂r

)= 1

c2

∂2v

∂t2, 0 < r < a, 0 < t, (1)

v(a, t) = 0, 0 < t, (2)


v(r,0) = f (r), 0 < r < a, (3)

∂v

∂t(r,0) = g(r), 0 < r < a. (4)

We start immediately with separation of variables, assuming v(r, t) =φ(r)T(t). The differential equation (1) becomes

1

r(rφ′)′T = 1

c2φT ′′,

and the variables may be separated by dividing by φT. Then we find

(rφ′(r))′

rφ(r)= T ′′(t)

c2T(t).

The two sides must both be equal to a constant (say, −λ2), yielding two linked,ordinary differential equations

T ′′ + λ2c2T = 0, 0 < t, (5)

(rφ′)′ + λ2rφ = 0, 0 < r < a. (6)

The boundary condition Eq. (2) is satisfied if

φ(a) = 0. (7)

Of course, because r = 0 is a singular point of the differential equation (6), weadd the requirement ∣∣φ(r)

∣∣ bounded at r = 0, (8)

which is equivalent to requiring that |v(r, t)| be bounded at r = 0. Werecognize that Eq. (6) is Bessel’s equation, of which the function φ(r)=J0(λr) is the solution bounded at r = 0. In order to satisfy the boundary con-dition Eq. (7), we must have

J0(λa) = 0,

which implies that

λn = αn

a, n = 1,2, . . . , (9)

where αn are the zeros of the function J0. Thus the eigenfunctions and eigen-values of Eqs. (6)–(8) are

φn(r) = J0(λnr), λ2n =

(αn

a

)2

.


The rest of our problem can now be dispatched easily. Returning to Eq. (5),we see that

Tn(t) = an cos(λnct) + bn sin(λnct),

and then for each n = 1,2, . . . we have a solution of Eqs. (1), (2), and (8):

vn(r, t) = φn(r)Tn(t).

The most general linear combination of the vn would be

v(r, t) =∞∑

n=1

J0(λnr)[an cos(λnct) + bn sin(λnct)

]. (10)

The initial conditions Eqs. (3) and (4) are satisfied if

v(r,0) =∞∑

n=1

an J0(λnr) = f (r), 0 < r < a,

∂v

∂t(r,0) =

∞∑n=1

bnλnc J0(λnr) = g(r), 0 < r < a.

As in the preceding section, the coefficients of these series are to be foundthrough the integral formulas

an = 1

Dn

∫ a

0f (r)J0(λnr)r dr, bn = 1

λncDn

∫ a

0g(r)J0(λnr)r dr,

Dn =∫ a

0

[J0(λnr)

]2r dr.

With the coefficients determined by these formulas, the function given inEq. (10) is the solution to the vibrating membrane problem that we startedwith.

General VibrationsHaving seen the simplest case of the vibrations of a circular membrane, wereturn to the more general case. The full problem was

1

r

∂

∂r

(r∂u

∂r

)+ 1

r2

∂2u

∂θ2= 1

c2

∂2u

∂t2, 0 < r < a, 0 < t. (11)

u(a, θ, t) = 0, 0 < t, (12)∣∣u(0, θ, t)∣∣ bounded, 0 < t, (13)

u(r, θ + 2π, t) = u(r, θ, t), 0 < r < a, 0 < t, (14)


u(r, θ,0) = f (r, θ), 0 < r < a, (15)

∂u

∂t(r, θ,0) = g(r, θ), 0 < r < a. (16)

Following the procedure suggested in Section 5.4, we assume that u has theproduct form

u = φ(r, θ)T(t)

and we find that Eq. (11) separates into two linked equations:

T ′′ + λ2c2T = 0, 0 < t, (17)

1

r

∂

∂r

(r∂φ

∂r

)+ 1

r2

∂2φ

∂θ2= −λ2φ, 0 < r < a. (18)

If we separate variables of the function φ by assuming φ(r, θ) = R(r)Q(θ),Eq. (18) takes the form

1

r

(rR′)′

Q + 1

r2RQ′′ = −λ2RQ.

The variables will separate if we multiply by r2 and divide by RQ. Then thepreceding equation may be put in the form

r(rR′)′

R+ λ2r2 = −Q′′

Q= µ2.

Finally we obtain two problems for R and Q:

Q′′ + µ2Q = 0, −π < θ ≤ π, (19a)

Q(θ + 2π) = Q(θ), (19b)

(rR′)′ − µ2

rR + λ2rR = 0, 0 < r < a, (20)∣∣R(0)

∣∣ bounded,

R(a) = 0.

As we observed before, the problem (19) has the solutions

µ20 = 0, Q0 = 1,

µ2m = m2, Qm = cos(mθ) and sin(mθ), m = 1,2,3, . . . .

Also, the differential equation (20) will be recognized as Bessel’s equation, thegeneral solution of which is (using µ = m)

R(r) = CJm(λr) + DYm(λr).


In order for the boundedness condition in Eq. (20) to be fulfilled, D must bezero. Then we are left with

R(r) = Jm(λr).

(Because any multiple of a solution is another solution, we can drop the con-stant C.) The boundary condition of Eq. (20) becomes

R(a) = Jm(λa) = 0,

implying that λa must be a root of the equation

Jm(α) = 0.

(See Table 1.) For each fixed integer m, αm1, αm2, αm3, . . . are the first, second,third, . . . solutions of the preceding equation. The values of λ for which Jm(λr)solves the differential equation and satisfies the boundary condition are

λmn = αmn

a, m = 0,1,2, . . . , n = 1,2,3, . . . .

Now that the functions R and Q are determined, we can construct φ. Form = 1,2,3, . . . and n = 1,2,3, . . . , both of the functions

Jm(λmnr) cos(mθ), Jm(λmnr) sin(mθ) (21)

are solutions of the problem Eq. (18), both corresponding to the same eigen-value λ2

mn. For m = 0 and n = 1,2,3, . . . , we have the functions

J0(λ0nr), (22)

which correspond to the eigenvalues λ20n. (Compare with the simple case.) The

function T(t) that is a solution of Eq. (17) is any combination of cos(λmnct)and sin(λmnct).

Now the solutions of Eqs. (11)–(14) have any of the forms

Jm(λmnr) cos(mθ) cos(λmnct), Jm(λmnr) sin(mθ) cos(λmnct),(23)

Jm(λmnr) cos(mθ) sin(λmnct), Jm(λmnr) sin(mθ) sin(λmnct)

for m = 1,2,3, . . . and n = 1,2,3, . . . . In addition, there is the special casem = 0, for which solutions have the form

J0(λ0nr) cos(λ0nct), J0(λ0nr) sin(λ0nct). (24)

The CD shows a few of these “standing waves” animated.The general solution of the problem Eqs. (11)–(14) will thus have the form

of a linear combination of the solutions in Eqs. (23) and (24). We shall use


several series to form the combination:

u(r, θ, t) =∑

n

a0n J0(λ0nr) cos(λ0nct)

+∑m,n

amn Jm(λmnr) cos(mθ) cos(λmnct)

+∑m,n

bmn Jm(λmnr) sin(mθ) cos(λmnct)

+∑

n

A0n J0(λ0nr) sin(λ0nct)

+∑m,n

Amn Jm(λmnr) cos(mθ) sin(λmnct)

+∑m,n

Bmn Jm(λmnr) sin(mθ) sin(λmnct). (25)

When t = 0, the last three sums disappear, and the cosines of t in the first threesums are all equal to 1. Thus

u(r, θ,0) =∑

n

a0n J0(λ0nr) +∑m,n

amn Jm(λmnr) cos(mθ)

+∑m,n

bmn Jm(λmnr) sin(mθ)

= f (r, θ), 0 < r < a, −π < θ ≤ π. (26)

We expect to fulfill this equality by choosing the a’s and b’s according tosome orthogonality principle. Since each function present in the series is aneigenfunction of the problem

∇2φ = −λ2φ, 0 < r < a,φ(a, θ) = 0,

φ(r, θ + 2π) = φ(r, θ), 0 < r < a,

we expect it to be orthogonal to each of the others (see Section 5.4, Exercise 7).This is indeed true: Any function from one series is orthogonal to all of thefunctions in the other series and also to the rest of the functions in its ownseries. To illustrate this orthogonality, we have∫∫

RJ0(λ0nr)Jm(λmnr) cos(mθ)dA

=∫ a

0J0(λ0nr)Jm(λmnr)

∫ π

−π

cos(mθ)dθr dr = 0, m �= 0. (27)


There are two other relations like this one involving functions from two dif-ferent series.

We already know that the functions within the first series are orthogonal toeach other: ∫ π

−π

∫ a

0J0(λ0nr)J0(λ0qr)r dr dθ = 0, n �= q.

Within the second series we must show that, if m �= p or n �= q, then

0 =∫ π

−π

∫ a

0Jm(λmnr) cos(mθ)Jp(λpqr) cos(pθ)r dr dθ. (28)

(Recall that r dr dθ = dA in polar coordinates.) Integrating with respect to θ

first, we see that the integral must be zero if m �= p, by the orthogonality ofcos(mθ) and cos(pθ). If m = p, the preceding integral becomes

π

∫ a

0Jm(λmnr)Jm(λmqr)r dr

after the integration with respect to θ . Finally, if n �= q, this integral is zero; thedemonstration follows the same lines as the usual Sturm–Liouville proof. (SeeSection 2.7.) Thus the functions within the second series are shown orthogonalto each other. For the functions of the last series, the proof of orthogonality issimilar.

Equipped now with an orthogonality relation, we can determine formulasfor the a’s and b’s. For instance,

a0n =∫ π

−π

∫ a0 f (r, θ)J0(λ0nr)r dr dθ

2π∫ a

0 J20(λ0nr)r dr

. (29)

The A’s and B’s are calculated from the second initial condition.It should now be clear that, while the computation of the solution to the

original problem is possible in theory, it will be very painful in practice. Worseyet, the final form of the solution Eq. (25) does not give a clear idea of what ulooks like. All is not wasted, however. We can say, from an examination of theλ’s, that the tone produced is not musical — that is, u is not periodic in t. Alsowe can sketch some of the fundamental modes of vibration of the membranecorresponding to some low eigenvalues (Fig. 9). The curves represent pointsfor which displacement is zero in that mode (nodal curves).

E X E R C I S E S

1. Verify that each of the functions in the series in Eq. (10) satisfies Eqs. (1),(2), and (8).

2. Derive the formulas for the a’s and b’s of Eq. (10).


Figure 9 Nodal curves: The curves in these graphs represent solutions ofφmn(r, θ) = 0. Adjacent regions bulge up or down, according to the sign. Onlythose φ’s containing the factor cos(mθ) have been used. (See the cover photo-graph.)

3. List the five lowest frequencies of vibration of a circular membrane.

4. Sketch the function J0(λnr) for n = 1,2,3.

5. What boundary conditions must the function φ of Eq. (18) satisfy?

6. Justify the derivation of Eqs. (19) and (20) from Eqs. (12)–(14) and (18).

7. Show that ∫ a

0Jm(λmnr)Jm(λmqr)r dr = 0, n �= q,

if

Jm(λmsa) = 0, s = 1,2, . . . .

8. Sketch the nodal curves of the eigenfunctions Eq. (21) corresponding toλ31, λ32, and λ33.

9. In the simple case of symmetric vibrations, we found the eigenfunctionsφ0n(r, θ) = J0(λ0nr), where J0(λ0na) = 0 for n = 1,2,3 . . . . The nodalcurves of φ03 are concentric circles. What are their radii (as multiples

5.8 Some Applications of Bessel Functions 329


of a)? What are the radii of the circles that are the nodal curves of φ0n(r, θ)

for general n?

10. The nodal curves of φmn(r, θ) are shown in Fig. 10.

a. By examining the figure, determine what values m and n have.

b. What is the numerical value of the eigenvalue λmn (as a multiple of a)for this eigenfunction?

c. What is the formula for the function φmn(r, θ) whose nodal curves areshown?

d. What is the frequency of vibration for the drumhead when it is vibrat-ing in this mode? (“In this mode” means “so that the displacement uequals a product solution in which this eigenfunction is a factor.”)

5.8 Some Applications of Bessel Functions

After the elementary functions, the Bessel functions are among the most usefulin engineering and physics. One reason for their usefulness is they solve a fairlygeneral differential equation. The general solution of

φ′′ + 1 − 2α

xφ′ +

[(λγ xγ−1

)2 − p2γ 2 − α2

x2

]φ = 0 (1)

is given by

φ(x) = xα[AJp(λxγ ) + BYp(λxγ )

].


Several problems in which the Bessel functions play an important role follow.The details of separation of variables, which should now be routine, are keptto a minimum.

A. Potential Equation in a CylinderThe steady-state temperature distribution in a circular cylinder with insulatedsurface is determined by the problem

1

r

∂

∂r

(r∂u

∂r

)+ ∂2u

∂z2= 0, 0 < r < a, 0 < z < b, (2)

∂u

∂r(a, z) = 0, 0 < z < b, (3)

u(r,0) = f (r), 0 < r < a, (4)

u(r,b) = g(r), 0 < r < a. (5)

Here we are considering the boundary conditions to be independent of θ , so uis independent of θ also.

Assuming that u = R(r)Z(z) we find that

(rR′)′ + λ2rR = 0, 0 < r < a, (6)

R′(a) = 0, (7)∣∣R(0)∣∣ bounded, (8)

Z′′ − λ2Z = 0. (9)

Condition (8) has been added because r = 0 is a singular point. The solutionof Eqs. (6)–(8) is

Rn(r) = J0(λnr), (10)

where the eigenvalues λ2n are defined by the solutions of

R′(a) = λ J ′0(λa) = 0. (11)

Because J ′0 = −J1, the λ’s are related to the zeros of J1. The first three eigen-

values are 0, (3.832/a)2, and (7.016/a)2. Note that R(0) = J0(0) = 1.The solution of the problem Eqs. (2)–(5) may be put in the form

u(r, z) = a0 + b0z +∞∑

n=1

J0(λnr)

[an

sinh(λnz)

sinh(λnb)+ bn

sinh(λn(b − z))

sinh(λnb)

]. (12)

5.8 Some Applications of Bessel Functions 331

The a’s and b’s are determined from Eqs. (4) and (5) by using the orthogonalityrelation ∫ a

0J0(λnr)J0(λmr)r dr = 0, n �= m.

B. Spherical WavesIn spherical (ρ, θ,φ) coordinates (see Section 5.9), the Laplacian operator ∇2

becomes

∇2u = 1

ρ2

∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

ρ2 sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)+ 1

ρ2 sin2(φ)

∂2u

∂θ2.

Consider a wave problem in a sphere when the initial conditions depend onlyon the radial coordinate ρ:

1

ρ2

∂

∂ρ

(ρ2 ∂u

∂ρ

)= 1

c2

∂2u

∂t2, 0 < ρ < a, 0 < t, (13)

u(a, t) = 0, 0 < t, (14)

u(ρ,0) = f (ρ), 0 < ρ < a, (15)

∂u

∂t(ρ,0) = g(ρ), 0 < ρ < a. (16)

Assuming u(ρ, t) = R(ρ)T(t), we separate variables and find

T ′′ + λ2c2T = 0, (17)(ρ2R′)′ + λ2ρ2R = 0, 0 < ρ < a, (18)

R(a) = 0, (19)∣∣R(0)∣∣ bounded. (20)

Again, the condition (20) has been added because ρ = 0 is a singular point.Equation (18) may be put into the form

R′′ + 2

ρR′ + λ2R = 0,

and comparison with Eq. (1) shows that α = −1/2, γ = 1, and ρ = 1/2; thusthe general solution of Eq. (18) is

R(ρ) = ρ−1/2[A J1/2(λρ) + BY1/2(λρ)

].

We know that near ρ = 0,

J1/2(λρ) ∼ const × ρ1/2,

Y1/2(λρ) ∼ const × ρ−1/2.


Thus in order to satisfy Eq. (20), we must have B = 0. It is possible to showthat

J1/2(λρ) = 2

π

sin(λρ)√λρ

, Y1/2(λρ) = − 2

π

cos(λρ)√λρ

.

Our solution to Eqs. (18) and (20) is, therefore,

R(ρ) = sin(λρ)

ρ, (21)

and Eq. (19) is satisfied if λ2n = (nπ/a)2. The solution of the problem of

Eqs. (13)–(16) can be written in the form

u(ρ, t) =∞∑

n=1

sin(λnρ)

ρ

[an cos(λnct) + bn sin(λnct)

]. (22)

The a’s and b’s are, as usual, chosen so that the initial conditions Eqs. (15)and (16) are satisfied.

C. Pressure in a BearingThe pressure in the lubricant inside a plane-pad bearing satisfies the problem

∂

∂x

(x3 ∂p

∂x

)+ x3 ∂2p

∂y2= −1, a < x < b, −c < y < c, (23)

p(a, y) = 0, p(b, y) = 0, −c < y < c, (24)

p(x,−c) = 0, p(x, c) = 0, a < x < b. (25)

(Here a and c are positive constants and b = a + 1.) Equation (23) is ellip-tic and nonhomogeneous. To reduce this equation to a more familiar one, letp(x, y) = v(x) + u(x, y), where v(x) satisfies the problem

(x3v′)′ = −1, a < x < b, (26)

v(a) = 0, v(b) = 0. (27)

Then, when v is found, u must be the solution of the problem

∂

∂x

(x3 ∂u

∂x

)+ x3 ∂2u

∂y2= 0, a < x < b, −c < y < c, (28)

u(a, y) = 0, u(b, y) = 0, −c < y < c, (29)

u(x,±c) = −v(x), a < x < b. (30)

If we now assume that u(x, y) = X(x)Y(y), the variables can be separated:

5.8 Some Applications of Bessel Functions 333(x3X′)′ + λ2x3X = 0, a < x < b, (31)

X(a) = 0, X(b) = 0, (32)

Y ′′ − λ2Y = 0, −c < y < c. (33)

Equation (31) may be put in the form

X′′ + 3

xX′ + λ2X = 0, a < x < b.

By comparing to Eq. (1) we find that α = −1, γ = 1, and p = 1 and that thegeneral solution of Eq. (31) is

X(x) = 1

x

(A J1(λx) + BY1(λx)

).

Because the point x = 0 is not included in the interval a < x < b, there is noproblem with boundedness. Instead we must satisfy the boundary conditionsEq. (32), which after some algebra have the form

A J1(λa) + BY1(λa) = 0,

A J1(λb) + BY1(λb) = 0.

Not both A and B may be zero, so the determinant of these simultaneousequations must be zero:

J1(λa)Y1(λb) − J1(λb)Y1(λa) = 0.

Some solutions of the equation are tabulated for various values of b/a. Forinstance, if b/a = 2.5, the first three eigenvalues λ2 are(

2.156

a

)2

,

(4.223

a

)2

,

(6.307

a

)2

.

We now can take Xn to be

Xn(x) = 1

x

(Y1(λna)J1(λnx) − J1(λna)Y1(λnx)

), (34)

and the solution of Eqs. (28)–(30) has the form

u(x, y) =∞∑

n=1

anXn(x)cosh(λny)

cosh(λnc). (35)

The a’s are chosen to satisfy the boundary conditions Eq. (30), using the or-thogonality principle∫ b

aXn(x)Xm(x)x3 dx = 0, n �= m.

Notice that Eqs. (31) and (32) make up a regular Sturm–Liouville problem.


E X E R C I S E S

1. Find the general solution of the differential equation

(xnφ′)′ + λ2xnφ = 0,

where n = 0,1,2, . . . .

2. Find the solution of the equation in Exercise 1 that is bounded at x = 0.

3. Find the solutions of Eq. (9), including the case λ2 = 0, and prove thatEq. (12) is a solution of Eqs. (2)–(4).

4. Show that any function of the form

u(ρ, t) = 1

ρ

(φ(ρ + ct) + ψ(ρ − ct)

)

is a solution of Eq. (13) if φ and ψ have at least two derivatives.

5. Find functions φ and ψ such that u(ρ, t) as given in Exercise 4 satisfiesEqs. (14)–(16).

6. Give the formula for the a’s and b’s in Eq. (12).

7. What is the orthogonality relation for the eigenfunctions of Eqs. (18)–(20)? Use it to find the a’s and b’s in Eq. (22).

8. Sketch the first few eigenfunctions of Eqs. (18)–(20).

9. Find the function v(x) that is the solution of Eqs. (26) and (27).

10. Use the technique of Example C to change the following problem into apotential problem:

∂2u

∂x2+ ∂2u

∂y2= −f (x), 0 < x < a, 0 < y < b,

u = 0 on all boundaries.

11. In Exercise 10, will the same technique work if f (x) is replaced by f (x, y)?

12. Verify that Eqs. (31) and (32) form a regular Sturm–Liouville problem.Show the eigenfunctions’ orthogonality by using the orthogonality of theBessel functions.

13. Find a formula for the an of Eq. (35).

14. Verify that Eq. (34) is a solution of Eqs. (28)–(30).

5.9 Spherical Coordinates; Legendre Polynomials 335

5.9 Spherical Coordinates;Legendre PolynomialsAfter the Cartesian and cylindrical coordinate systems, the one most fre-quently encountered is the spherical system (Fig. 11), in which

x = ρ sin(φ) cos(θ),

y = ρ sin(φ) sin(θ),

z = ρ cos(φ).

The variables are restricted by 0 ≤ ρ, 0 ≤ θ < 2π , 0 ≤ φ ≤ π . In this coordi-nate system the Laplacian operator is

∇2u = 1

ρ2

{∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)+ 1

sin2(φ)

∂2u

∂θ2

}.

From what we have seen in other cases, we expect solvable problems inspherical coordinates to reduce to one of the following.

Problem 1. ∇2u = −λ2u in R, plus homogeneous boundary conditions.

Problem 2. ∇2u = 0 in R, plus homogeneous boundary conditions on facingsides (where R is a generalized rectangle in spherical coordinates).

Problem 1 would come from a heat or wave equation after separating outthe time variable. Problem 2 is a part of the potential problem.

The complete solution of either of these problems is very complicated, but anumber of special cases are simple, important, and not uncommon. We havealready seen Problem 1 solved (Section 5.8) when u is a function of ρ only.A second important case is Problem 2, when u is independent of the variable θ .We shall state a complete boundary value problem and solve it by separation

Figure 11 Spherical coordinates.


of variables:

1

ρ2

{∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)}= 0,

0 < ρ < c, 0 < φ < π, (1)

u(c, φ) = f (φ), 0 < φ < π. (2)

From the assumption u(ρ,φ) = R(ρ)�(φ), it follows that

(ρ2R′(ρ))′

R(ρ)+ (sin(φ)�′(φ))′

sin(φ)�(φ)= 0.

Both terms are constant, and the second is negative, −µ2, because the bound-ary condition at ρ = c will have to be satisfied by a linear combination of func-tions of φ. The separated equations are(

ρ2R′)′ − µ2R = 0, 0 < ρ < c, (3)(sin(φ)�′)′ + µ2 sin(φ)� = 0, 0 < φ < π. (4)

Neither equation has a boundary condition. However, ρ = 0 is a singular pointof the first equation, and both φ = 0 and ρ = π are singular points of the sec-ond equation. (At these points, the coefficient of the highest-order derivativeis zero, while some other coefficient is nonzero.) At each of the singular points,we impose a boundedness condition:

R(0) bounded, �(0) and �(π) bounded.

Equation (4) can be simplified by the change of variables x = cos(φ),�(φ) = y(x). (Of course, x is not the Cartesian coordinate.) By the chain rule,the relevant derivatives are

d�

dφ= − sin(φ)

dy

dx,

d

dφ

(sin(φ)

d�

dφ

)= sin3(φ)

d2y

dx2− 2 sin(φ) cos(φ)

dy

dx.

The differential equation becomes

sin2(φ)d2y

dx2− 2 cos(φ)

dy

dx+ µ2y = 0,

or, in terms of x alone,(1 − x2

)y′′ − 2xy′ + µ2y = 0, −1 < x < 1. (5)

In addition, we require that y(x) be bounded at x = ±1.


Solutions of the differential equation are usually found by the power seriesmethod. Assume that y(x) = a0 + a1x + · · · + akxk + · · · . The terms of thedifferential equations are then

y′′ = 2a2 + 3 · 2a3x + 4 · 3a4x2 + · · · + (k + 2)(k + 1)ak+2xk + · · ·−x2y′′ = − 2a2x2 − · · · − k(k − 1)akxk + · · ·−2xy′ = − 2a1x − 2a2x2 − · · · − 2kakxk − · · ·

µ2y = µ2a0 + µ2a1x + µ2a2x2 + · · · + µ2akxk + · · · .When this tableau is added vertically, the left-hand side is zero, according to

the differential equation. The right-hand side adds up to a power series, eachof whose coefficients must be zero. We therefore obtain the following relations:

2a2 + µ2a0 = 0,

6a3 + (µ2 − 2

)a1 = 0,

(k + 2)(k + 1)ak+2 + [µ2 − k(k + 1)

]ak = 0.

The last equation actually includes the first two, apparently special, cases. Wemay write the general relation as

ak+2 = k(k + 1) − µ2

(k + 2)(k + 1)ak,

valid for k = 0,1,2, . . . .Suppose for the moment that µ2 is given. A short calculation gives the first

few coefficients:

a2 = −µ2

2a0, a3 = 2 − µ2

6a1,

a4 = 6 − µ2

12a2, a5 = 12 − µ2

20a3,

= 6 − µ2

12· −µ2

2a0, = 12 − µ2

20· 2 − µ2

6a1.

It is clear that all the a’s with even index will be multiples of a0 and those withodd index will be multiples of a1. Thus y(x) equals a0 times an even functionplus a1 times an odd function, with both a0 and a1 arbitrary.

It is not difficult to prove that odd and even series produced by this processdiverge at both x = ±1, for general µ2. However, when µ2 has one of the spe-cial values

µ2 = µ2n = n(n + 1), n = 0,1,2, . . . ,

one of the two series turns out to have all zero coefficients after an. For in-stance, if µ2 = 3 · 4, then a5 = 0, and all subsequent coefficients with odd


P0(x) = 1P1(x) = xP2(x) = (3x2 − 1)/2P3(x) = (5x3 − 3x)/2P4(x) = (35x4 − 30x2 + 3)/8

Table 3 Legendre polynomials

Figure 12 Graphs of the first five Legendre polynomials.

index are also zero. Hence, one of the solutions of

(1 − x2

)y′′ − 2xy′ + 12y = 0

is the polynomial a1(x − 5x3/3). The other solution is an even function un-bounded at both x = ±1.

Now we see that the boundedness conditions can be satisfied only if µ2 isone of the numbers 0,2,6, . . . ,n(n + 1), . . . . In such a case, one solution ofthe differential equation is a polynomial (naturally bounded at x = ±1). Whennormalized by the condition y(1) = 1, these are called Legendre polynomials,written Pn(x). Table 3 provides the first five Legendre polynomials. Figure 12shows their graphs.


Since the differential equation (5) is easily put into self-adjoint form,

((1 − x2)y′)′ + µ2y = 0, −1 < x < 1,

it is routine to show that the Legendre polynomials satisfy the orthogonalityrelation ∫ 1

−1Pn(x)Pm(x)dx = 0, n �= m.

By direct calculation, it can be shown that

∫ 1

−1P2

n(x)dx = 2

2n + 1. (6)

A compact way of representing the Legendre polynomials is by means of Ro-drigues’ formula,

Pn(x) = 1

n!2n

dn

dxn

[(x2 − 1)n

]. (7)

Elementary algebra and calculus show that the nth derivative of (x2 − 1)n isa polynomial of degree n. Substituting this polynomial into the differentialequation (5), with µ2 = n(n + 1), shows that it is a solution — bounded, ofcourse. Therefore, it is a multiple of the Legendre polynomial Pn(x). ThroughRodrigues’ formula or otherwise, it is possible to prove the following two for-mulas, which relate three consecutive Legendre polynomials:

(2n + 1)Pn(x) = P ′n+1(x) − P ′

n−1(x), (8)

(n + 1)Pn+1(x) + nPn−1(x) = (2n + 1)xPn(x). (9)

In order to use Legendre polynomials in boundary value problems, we needto be able to express a given function f (x) in the form of a Legendre series,

f (x) =∞∑

n=0

bnPn(x), −1 < x < 1.

From the orthogonality relation and the integral, Eq. (6), it follows that thecoefficient in the series must be

bn = 2n + 1

2

∫ 1

−1f (x)Pn(x)dx. (10)

The convergence theorem for Legendre series is analogous to the one forFourier series in Chapter 1.


Theorem. If f (x) is sectionally smooth on the interval −1 < x < 1, then at everypoint of that interval the Legendre series of f is convergent, and

∞∑n=0

bnPn(x) = f (x+) + f (x−)

2. �

From Eq. (10) for the coefficient of a Legendre series and from the fact thatthe Legendre polynomials are odd or even, we see that an odd function willhave only odd-indexed coefficients that are nonzero, and an even function willhave only even-indexed coefficients that are nonzero. Furthermore, if a func-tion f is given on the interval 0 < x < 1, then its odd and even extensions haveodd and even Legendre series, and f is represented by either in that interval:

f (x) =∑

n even

bnPn(x), 0 < x < 1,

bn = (2n + 1)

∫ 1

0f (x)Pn(x)dx (n even), (11)

f (x) =∑n odd

bnPn(x), 0 < x < 1,

bn = (2n + 1)

∫ 1

0f (x)Pn(x)dx (n odd). (12)

Because the Pn(x) are polynomials, the integral equation (10) for any spe-cific coefficient can be done in closed form for a variety of functions f (x).However, getting an as a function of n is not so easy. Fortunately, some ele-mentary integrals can be done using the differential equation(

(1 − x2)P ′n

)′ + n(n + 1)Pn = 0.

(1) First, separate the two terms of the differential equation and integrate:

n(n + 1)

∫Pn(x)dx =

∫−(

(1 − x2)P ′n

)′dx

= −(1 − x2

)P ′

n(x).

This equation may be solved for the integral if n �= 0.(2) Now multiply through the differential equation by x, separate terms, and

integrate:

n(n + 1)

∫xPn(x)dx =

∫−x

((1 − x2)P ′

n

)′dx

= −x(1 − x2

)P ′

n +∫ (

1 − x2)P ′

n dx

= −x(1 − x2

)P ′

n + (1 − x2

)Pn −

∫(−2x)Pn dx.


Next, move the last term to the left-hand member of the equation to find

(n + 2)(n − 1)

∫xPn(x)dx = (

1 − x2)(

Pn(x) − xP ′n(x)

).

This equation can be solved for the integral on the left, provided that n �= 1.(For n = 1, the integration is done directly.)

Summary ∫Pn(x)dx = −(1 − x2)

n(n + 1)P ′

n(x), (13)

∫xPn(x)dx = (1 − x2)

(n + 2)(n − 1)

(Pn(x) − xP ′

n(x))

(14)

These integration formulas are useful if we can evaluate Pn(x) and P ′n(x)

easily for any x. The relations in Eqs. (8) and (9) are useful for this purpose.We illustrate by finding Pn(0). First, note that Pn(0) = 0 for odd values of n,because the Legendre polynomials with odd index are odd functions of x. Forodd n, Eq. (9) gives

(n + 1)Pn+1(0) + nPn−1(0) = 0,

or

Pn+1(0) = − n

n + 1Pn−1(0).

Because P0(0) = 1, we find successively that

P2(0) = −1

2, P4(0) = 1 · 3

2 · 4, P6(0) = −1 · 3 · 5

2 · 4 · 6,

or in general

Pn(0) = (−1)n/2 1 · 3 · · · (n − 1)

2 · 4 · · ·n, n = 2,4,6, . . .

Pn(0) = 0, n = 1,3,5, . . . .

(15)

Similarly, but not as easily, Eq. (8) can be used to find the values of P ′n(0). It

is simpler to use the relation

P ′n(0) = nPn−1(0), (16)

which can be derived from Eqs. (8) and (9).


Example.Let

f (x) ={−1, −1 < x < 0,

1, 0 < x < 1.

The Legendre series will contain only odd-indexed polynomials, and their co-efficients are

bn = (2n + 1)

∫ 1

0Pn(x)dx (n odd)

= − 2n + 1

n(n + 1)

[(1 − x2

)P ′

n(x)]1

0

= 2n + 1

n(n + 1)P ′

n(0) = 2n + 1

n + 1Pn−1(0)

= (−1)(n−1)/2 1 · 3 · 5 · · · (n − 2)

2 · 4 · 6 · · · (n − 1)· 2n + 1

n + 1(n = 3,5,7, . . .).

Specifically we find b1 = 3/2 (by a separate calculation), b3 = −7/8, b5 =11/16, . . . . Because f (x) is indeed sectionally smooth,

f (x) = 3

2P1(x) − 7

8P3(x) + 11

16P5(x) − · · · .

See Fig. 13 for graphs of the partial sums of this series. �

(a) (b)

Figure 13 Graphs of a function and a partial sum of its Legendre series:(a) through P9(x) for the function f (x) in the example; (b) through P6(x) forf (x) = |x|, −1 < x < 1. Compare with the partial sums of the Fourier series, Figs. 9and 10 of Chapter 1.


Figure 14 The nodal curves of the zonal harmonics are the parallels(φ = constant) on a sphere, where Pn(cos(φ)) = 0. The nodal curves are shownin projection for n = 1,2,3,4. See the CD for color versions.

Summary

The solution of the eigenvalue problem((1 − x2)y′)′ + µ2y = 0, −1 < x < 1,

y(x) bounded at x = −1 and at x = 1,

is y(x) = Pn(x), µ2n = n(n + 1), n = 0,1,2, . . . .

The solution of the eigenvalue problem(sin(φ)�′)′ + µ2 sin(φ)� = 0,

�(φ) bounded at φ = 0 and at φ = π,

is �(φ) = Pn(cos(φ)), µ2n = n(n + 1), n = 0,1,2, . . . .

The Legendre polynomials Pn(cos(φ)) are often called zonal harmonics be-cause their nodal lines (loci of solutions of Pn(cos(φ)) = 0) divide a sphereinto zones, as shown in Fig. 14.

E X E R C I S E S

1. Equation (4) may be solved by assuming

�(φ) = 1

2a0 +

∞∑1

ak cos(kφ).

Find the relations among the coefficients ak by computing the terms of theequation in the form of series. Use the identities

sin(φ) sin(kφ) = 1

2

[cos

((k − 1)φ

) − cos((k + 1)φ

)],

sin(φ) cos(kφ) = −1

2

[sin

((k − 1)φ

) + sin((k + 1)φ

)].


Show that the coefficients are all zero after an if µ2 = n(n + 1).

2. Derive the formula for the coefficients bn, as shown in Eq. (10).

3. Find P5(x), first from the formulas for the a’s and second by using Eq. (9)with n = 4.

4. Verify Eqs. (6) and (7) for n = 0,1,2 and Eq. (9) for n = 2,3.

5. One of the solutions of (1 − x2)y′′ − 2xy′ = 0 is y(x) = 1(µ2 = 0). Findanother independent solution of this differential equation.

6. Show that the orthogonality relation for the eigenfunctions �n(φ) =Pn(cos(φ)) is ∫ π

0�n(φ)�m(φ) sin(φ)dφ = 0, n �= m.

7. Obtain the relation

P ′n+1(x) = (n + 1)Pn(x) + xP ′

n(x)

by differentiating Eq. (9) and eliminating P ′n−1 between that and Eq. (8).

Note that Eq. (16) follows from this relation.

8. Let F = (x2 − 1)n. Show that F satisfies the differential equation(x2 − 1

)F′ = 2nxF.

9. Differentiate both sides of the preceding equation n+1 times to show thatthe nth derivative of F satisfies Legendre’s equation (5). Use Leibniz’s rulefor derivatives of a product.

10. Obtain Eq. (6) by these manipulations:

a. Multiply through Eq. (9) by Pn+1, integrate from −1 to 1, and use theorthogonality of Pn+1 with Pn−1.

b. Replace (2n + 1)Pn by means of Eq. (8).

c. Pn+1 is orthogonal to xP ′n−1, which is a polynomial of degree n.

d. Solve what remains for the desired integral.

11. Find the Legendre series for the function f (x) = |x|, −1 < x < 1.

12. Find the Legendre series for the following function. Note that f (x) − 1/2is an odd function.

f (x) ={0, −1 < x < 0,

1, 0 < x < 1.

5.10 Some Applications of Legendre Polynomials 345

5.10 Some Applications of LegendrePolynomialsIn this section we follow through the details involved in solving some problemsin which Legendre polynomials are used. First, we complete the problem statedin the previous section.

A. Potential in a SphereWe consider the axially symmetric potential equation — that is, with no vari-ation in the longitudinal- or θ -direction. The unknown function u might rep-resent an electrostatic potential, steady-state temperature, etc.

1

ρ2

{∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)}= 0,

0 < ρ < c, 0 < φ < π, (1)

u(c, φ) = f (φ), 0 < φ < π. (2)

Of course, the function u is to be bounded at the singular points φ = 0, φ = π ,and ρ = 0. The assumption that u has the product form, u(ρ,φ) = �(φ)R(ρ),allows us to transform the partial differential equation into

(ρ2R′(r))′

R(r)+ (sin(φ)�′(φ))′

sin(φ)�(φ)= 0.

From here we obtain equations for R and � individually,(ρ2R′)′ − µ2R = 0, 0 < ρ < c, (3)(sin(φ)�′)′ + µ2 sin(φ)� = 0, 0 < φ < π. (4)

In Section 5.9 we found the eigenfunctions of Eq. (4), subject to the bound-edness conditions at φ = 0 and π , to be �n(φ) = Pn(cos(φ)), correspondingto the eigenvalues µ2

n = n(n + 1). We must still solve Eq. (3) for R. After thedifferentiation has been carried out, the problem for R becomes

ρ2R′′n + 2ρR′

n − n(n + 1)Rn = 0, 0 < ρ < c,

Rn bounded at ρ = 0.

The equation is of the Cauchy–Euler type, solved by assuming R = ρα anddetermining α. Two solutions, ρn and ρ−(n+1), are found, of which the sec-ond is unbounded at ρ = 0. Hence Rn = ρn, and our product solutions of thepotential equation have the form

un(ρ,φ) = Rn(ρ)�n(φ) = ρnPn

(cos(φ)

).


The general solution of the partial differential equation that is bounded in theregion 0 < ρ < c, 0 < φ < π is thus the linear combination

u(ρ,φ) =∞∑

n=0

bnρnPn

(cos(φ)

). (5)

At ρ = c, the boundary condition becomes

u(c, φ) =∞∑

n=0

bncnPn

(cos(φ)

) = f (φ), 0 < φ < π. (6)

The coefficients bn are then found to be

bn = 2n + 1

2cn

∫ π

0f (φ)Pn

(cos(φ)

)sin(φ)dφ. (7)

B. Heat Equation on a Spherical ShellThe temperature on a spherical shell satisfies the three-dimensional heat equa-tion. If initially there is no dependence on θ , then there will never be such de-pendence. Furthermore, if the shell is thin (thickness much less than averageradius R), we may also assume that temperature does not vary in the radialdirection. The heat equation then becomes one-dimensional:

1

sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)= R2

k

∂u

∂t, 0 < φ < π, 0 < t, (8)

u(φ,0) = f (φ), 0 < φ < π. (9)

Naturally, we require boundedness of u at φ = 0 and φ = π .The assumption of a product form for the solution, u(φ, t) = �(φ)T(t),

leads to the conclusion that

(sin(φ)�′(φ))′

sin(�(φ))= R2T ′(t)

kT(t)= −µ2.

Thus, we have the eigenvalue problem(sin(φ)�′)′ + µ2 sin(φ)� = 0, 0 < φ < π,

�(0) and �(π) bounded.

The solution of this problem was found in Section 5.9 to be µ2 = n(n + 1) and

�n(φ) = Pn

(cos(φ)

), n = 0,1,2, . . . .

Obviously, the other factor in a product solution must be

Tn(t) = exp(−n(n + 1)kt/R2

).


Now, a series of constant multiples of product solutions is the most generalsolution of our problem:

u(φ, t) =∞∑

n=0

bnPn

(cos(φ)

)e−n(n+1)kt/R2

. (10)

The initial condition, Eq. (9), now takes the form of a Legendre series,

∞∑n=0

bnPn

(cos(φ)

) = f (φ), 0 < φ < π. (11)

From the information in Section 5.9, we know that the coefficients bn must bechosen to be

bn = 2n + 1

2

∫ π

0Pn

(cos(φ)

)f (φ) sin(φ)dφ.

Then if f (φ) is sectionally smooth for 0 < φ < π , the series of Eq. (11) actuallyequals f (φ), and thus the function u(φ, t) in Eq. (10) satisfies the problemoriginally posed.

For instance, if f (φ) = T0 in the northern hemisphere (0 < φ < π/2) andf (φ) = −T0 in the southern (π/2 < φ < π), then the coefficients are

bn = 2n + 1

2

∫ π

0f (φ)Pn

(cos(φ)

)sin(φ)dφ

= 2n + 1

2

[∫ 1

−1f(cos−1(x)

)Pn(x)dx

]

= T02n + 1

n + 1Pn−1(0),

as found in the previous section. Figure 15 shows graphs of u(φ, t) as a func-tion of φ in the interval 0 < φ < π for various times. The CD shows an ani-mated version of the solution.

C. Spherical WavesIn Section 5.8, we solved the wave equation in spherical coordinates for thecase where the initial conditions depend only on the radial variable ρ. Now weconsider the case where the variable φ is also present. A full statement of the


Figure 15 Graphs of the solution of the example problem, with u(φ,0) positivein the north and negative in the south. The function u(φ, t) is shown as a functionof φ in the range 0 to π for times chosen so that the dimensionless time kt/R2

takes the values 0, 0.01, 0.1, and 1; for convenience, T0 = 100.

problem is

1

ρ2

∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

ρ2 sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)= 1

c2

∂2u

∂t2,

0 < ρ < a, 0 < φ < π, 0 < t,

u(a, φ, t) = 0, 0 < φ < π, 0 < t,

u(ρ,φ,0) = f (ρ,φ), 0 < ρ < a, 0 < φ < π,

∂u

∂t(ρ,φ,0) = g(ρ,φ), 0 < ρ < a, 0 < φ < π.

(12)

As usual, we require in addition that u be bounded as ρ → 0 and as φ → 0and φ → π .

First, we seek solutions in the product form u(ρ,φ, t) = R(ρ)�(φ)T(t).Inserting u in this form into the partial differential equation (12) and manip-ulating, we find

1

ρ2

((ρ2R′)′

R+ (sin(φ)�′)′

sin(φ)�

)= T ′′

c2T. (13)

Both sides of this equation must have the same and constant value, say, −λ2.Thus, we must have

(ρ2R′)′

R+ (sin(φ)�′)′

sin(φ)�= −λ2ρ2.


Again we see that the ratio containing � must be constant, say, −µ2. Hence,we have two separate problems for the functions � and R:(

sin(φ)�′)′ + µ2 sin(φ)� = 0, 0 < φ < π,

�(φ) bounded at φ = 0, π,(ρ2R′)′ − µ2R + λ2ρ2R = 0, 0 < ρ < a,

R(a) = 0,

R(ρ) bounded at 0.

The first of these problems is now quite familiar, and we know its solution tobe

µ2n = n(n + 1), �n(φ) = Pn

(cos(φ)

), n = 0,1,2, . . . .

The second problem is less familiar. In standard form, the differential equa-tion is

R′′ + 2

ρR′ − µ2

ρ2R + λ2R = 0.

Comparison with the four-parameter form of Bessel’s equation (Eq. (1) of Sec-tion 5.8) shows α = −1/2, γ = 1, and p2 = µ2 + α2. Since µ = n(n + 1),p2 = n2 + n + 1

4 , and then p = n + 12 . Thus, the general solution of the differ-

ential equation is

Rn(ρ) = ρ−1/2[A Jn+1/2(λρ) + BYn+1/2(λρ)

].

The fact that the Bessel functions of the second kind, Yp(λρ), are un-bounded at ρ = 0 allows us to discard them from the solution, leaving

Rn(ρ) = ρ−1/2Jn+1/2(λρ)

as the bounded solution. These functions occur frequently in problems inspherical coordinates. Sometimes the functions

jn(z) =√

π

2zJn+1/2(z),

called spherical Bessel functions of the first kind of order n, are introduced. Asnoted in Section 5.8, there is a relation to sines and cosines:

j0(z) = sin(z)/z,

j1(z) = (sin(z) − z cos(z)

)/z2,

j2(z) = ((3 − z2) sin(z) − 3z cos(z)

)/z3.

We have yet to satisfy the boundary condition Rn(a) = 0. This cannot bedone by formula, except for n = 0. In this case, R0(a) = 0 comes down to


sin(λa)/λa = 0, so λm = mπ/a for m = 1,2, . . . . For other n’s, solutions ofJn+1/2(λa) = 0 must be found numerically. For instance, for n = 1 the equationis

sin(λa) − λa cos(λa) = 0,

with solutions λa = 4.493, 7.725, 10.904, . . . . (See Handbook of MathematicalFunctions by Abramowitz and Stegun, listed in the Bibliography.)

Finally we can put together some product solutions. Clearly the factor func-tion T(t) will be a sine or cosine of λct. Thus our product solutions have theform

ρ−1/2Jn+1/2(λnmρ)Pn

(cos(φ)

)sin(λnmct),

ρ−1/2Jn+1/2(λnmρ)Pn

(cos(φ)

)cos(λnmct).

The solution u(ρ,φ, t) will be an infinite series of constant multiples of thesefunctions. We will not write it out.

Let us summarize some of the information we have obtained. First, the fre-quencies of vibration of a sphere are λmnc (radians per unit time), where λmn

is the mth positive solution of

Jn+1/2(λa) = 0.

Second, the nodal surfaces (loci of points where a product solution is 0 for alltime) are the values of ρ and φ for which

Jn+1/2(λmnρ)Pn

(cos(φ)

) = 0.

One or the other factor must be 0, so these surfaces are either concentricspheres, ρ = const., determined by Jn+1/2(λmnρ) = 0, or else cones φ = const.,determined by Pn(cos(φ)) = 0.

Finally, let us observe that, because P0(cos(φ)) ≡ 1, the product solutionswith n = 0 are precisely what we found as product solutions of the problem inSection 5.8, Part B.

E X E R C I S E S

1. Solve the potential equation in the sphere 0 < ρ < 1, 0 < φ < π with theboundary condition

u(1, φ) ={

1, 0 < φ < π/2,0, π/2 < φ < π ,

together with appropriate boundedness conditions.


2. Solve the potential equation in a hemisphere, 0 < ρ < 1, 0 < φ < π/2,subject to boundedness conditions at ρ = 0 and φ = 0, and the boundaryconditions

u(1, φ) = 1, 0 < φ < π/2,

u(ρ,π/2) = 0, 0 < ρ < 1.

Hint: Use odd-order Legendre polynomials.

3. Solve this heat problem with convection on a spherical shell of radius R:

1

R2 sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)− γ 2(u − T) = 1

k

∂u

∂t,

0 < φ < π, 0 < t,

u(φ,0) = 0, 0 < φ < π.

Think carefully about the physical situation before attempting a solution.

4. Solve this heat problem on a hemispherical shell of radius R:

1

R2 sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)= 1

k

∂u

∂t, 0 < φ < π/2, 0 < t,

∂u

∂φ(π/2, t) = 0, 0 < t,

u(φ,0) = cos(φ), 0 < φ < π/2.

5. Solve the eigenvalue problem

1

ρ2

[∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)]= −λ2u,

0 < ρ < a, 0 < φ < π/2,

u(a, φ) = 0, 0 < φ < π/2,

u(r,π/2) = 0, 0 < r < a,

subject to boundedness conditions at ρ = 0 and at φ = 0.

6. In Part C of this section we mention nodal surfaces (i.e., surfaces wherethe function is 0). Find the nodal surfaces of the function

ρ−1/2J3/2(λρ)P1

(cos(φ)

)if λ is the second positive solution of J3/2(λ) = 0.

7. Describe in words the nodal surfaces for

ρ−1/2J5/2(λρ)P2

(cos(φ)

)


if λ is the second positive solution of J5/2(λ) = 0.

8. Solve the potential problem in the exterior of a sphere.

1

ρ2

[∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)]= 0,

R < ρ, 0 < φ < π,

u(R, φ) = f (φ), 0 < φ < π.

9. L.M. Chiappetta and D.R. Sobel [Temperature distribution within a hemi-sphere exposed to a hot gas stream, SIAM Review, 26 (1984): 575–577]analyze the steady-state temperature in the rounded tip of a combustion-gas sampling probe. The tip is approximately hemispherical in shape. Itsouter surface is exposed to hot gases at temperature TG, and its baseis cooled by water at temperature TW circulating inside the probe. IfT(ρ,φ) is the temperature inside the tip, it should satisfy the condi-tions

1

ρ2

[∂

∂ρ

(ρ2 ∂T

∂ρ

)+ 1

sin(φ)

∂

∂φ

(sin(φ)

∂T

∂φ

)]= 0,

0 < ρ < R, 0 < φ <π

2,

T(ρ,π/2) = TW , 0 < ρ < R,

k∂T

∂ρ(R, φ) = h

[TG − T(R, φ)

], 0 < φ <

π

2

together with boundedness conditions at ρ = 0 and at φ = 0.

The authors then change the variables to simplify the problem. Letr = ρ/R, u(r, φ) = T(ρ,φ) − TW , and show that the problem for u be-comes

∂

∂ρ

(ρ2 ∂u

∂ρ

)+ 1

sin(φ)

∂

∂φ

(sin(φ)

∂u

∂φ

)= 0, 0 < r < 1, 0 < φ < π/2,

u(r,π/2) = 0, 0 < r < 1,

K∂u

∂r(1, φ) + u(1, φ) = D, 0 < φ < π/2,

where K = k/hR and D = TG − TW .

10. Solve the problem in Exercise 9. Hint: Use odd-indexed Legendre polyno-mials to satisfy the boundary condition at φ = π/2.



We have seen just a few problems in two or three dimensions, but they aresufficient to illustrate the complications that may arise. A serious drawbackto the solution by separation of variables is that double and triple series tendto converge slowly, if at all. Thus, if a numerical solution to a two- or three-dimensional problem is needed, it may be advisable to sidestep the analyticalsolution by using an approximate numerical technique from the beginning.

One advantage of using special coordinate systems is that some problemsthat are two-dimensional in Cartesian coordinates may be one-dimensionalin another system. This is the case, for instance, when distance from a point(r in polar or ρ in spherical coordinates) is the only significant space variable.Of course, nonrectangular systems may arise naturally from the geometry of aproblem.

As Sections 5.3 and 5.4 point out, solving the two-dimensional heat or waveequation in a region R of the plane depends on being able to solve the eigen-value problem ∇2φ = −λ2φ in R with φ = 0 on the boundary. The solutionof this problem in a region bounded by coordinate curves (that is, in a gen-eralized rectangle) is known for many coordinate systems. We have discussedthe most common cases; others can be found in Methods of Theoretical Physicsby Morse and Feshbach. Information about the special functions involved isavailable from the Handbook of Mathematical Functions by Abramowitz andStegun and also from Special Functions of Mathematics for Engineers by L.C.Andrews. Eigenfunctions and eigenvalues are known for a few regions that arenot generalized rectangles. (See Miscellaneous Exercises 20 and 21 in the textthat follows.)

Eigenvalues of the Laplacian in a region can be estimated by a Rayleigh quo-tient, much as in Section 3.5. Furthermore, we have theorems of the follow-ing type. Let λ2

1 be the lowest eigenvalue of ∇φ = −λ2φ in R with φ = 0on the boundary. Let λ2

1 have the same meaning for another region, R. If Rfits inside R, then λ2

1 ≥ λ21. [The smaller the region, the larger the first eigen-

value. For further information, see Methods of Mathematical Physics, Vol. 1,by Courant and Hilbert. In the famous article “Can one hear the shape of adrum?,” American Mathematical Monthly, 73 (1966): 1–23], Mark Kac showsthat one can find the area, perimeter, and connectivity of a region from theeigenvalues of the Laplacian for that region. However, Kac’s title question hasbeen answered negatively. In the Bulletin of the American Mathematical Soci-ety, 27 (1992): 134–138, authors C. Gordon, D.L. Webb, and S. Wolpert displaytwo plane regions, or “drums,” of different shapes, on which the Laplacian hasexactly the same eigenvalues.

The nodal curves of a membrane shown in Fig. 9 can be realized physically.Photographs of such curves, along with an explanation of the physics of the


vibrating membrane, will be found in The Physics of Musical Instruments, byFletcher and Rossing.

Chapter ReviewSee the CD for review questions and special exercises.

Miscellaneous Exercises1. Solve the heat problem

∇2u = 1

k

∂u

∂t, 0 < x < a, 0 < y < b, 0 < t,

∂u

∂x(0, y, t) = d0,

∂u

∂x(a, y, t) = 0, 0 < y < b, 0 < t,

u(x,0, t) = 0, u(x,b, t) = 0, 0 < x < a, 0 < t,

u(x, y,0) = Tx

a, 0 < x < a, 0 < y < b.

2. Same as Exercise 1, but the initial condition is

u(x, y,0) = Ty

b, 0 < x < a, 0 < y < b.

3. Let u(x, y, t) be the solution of the heat equation in a rectangle as statedhere. Find an expression for u(a/2,b/2, t). Write out the first threenonzero terms for the case a = b.

∇2u = 1

k

∂u

∂t, 0 < x < a, 0 < y < b, 0 < t,

u = 0 on all boundaries,u(x, y,0) = T, 0 < x < a, 0 < y < b.

4. Find the nodal lines of the square membrane. These are loci of pointssatisfying φmn(x, y) = 0, where φmn satisfies ∇2φ = −λ2φ in the squareand φ = 0 on the boundary.

5. Find the solution of the boundary value problem

1

r

d

dr

(r

du

dr

)= −1, 0 < r < a,

u(0) bounded, u(a) = 0,


both directly and by assuming that both u(r) and the constant function 1have Bessel series on the interval 0 < r < a:

u(r) =∞∑

n=1

Cn J0(λnr), 0 < r < a,

1 =∞∑

n=1

cn J0(λnr), 0 < r < a.(Hint:

1

r

d

dr

(r

d J0(λr)

dr

)= −λ2J0(λr).

)

6. Suppose that w(x, t) and v(y, t) are solutions of the partial differentialequations

∂2w

∂x2= 1

k

∂w

∂t,

∂2v

∂y2= 1

k

∂v

∂t.

Show that u(x, y, t) = w(x, t)v(y, t) satisfies the two-dimensional heatequation

∂2u

∂x2+ ∂2u

∂y2= 1

k

∂u

∂t.

7. Use the idea of Exercise 6 to solve the problem stated in Exercise 1.

8. Let w(x, y) and v(z, t) satisfy the equations

∂2w

∂x2+ ∂2w

∂y2= 0,

∂2v

∂z2= 1

c2

∂2v

∂t2.

Show that the product u(x, y, z, t) = w(x, y)v(z, t) satisfies the three-dimensional wave equation

∂2u

∂x2+ ∂2u

∂y2+ ∂2u

∂z2= 1

c2

∂2u

∂t2.

9. Find the product solutions of the equation

1

r

∂

∂r

(r∂u

∂r

)= 1

k

∂u

∂t, 0 < r, 0 < t,

that are bounded as r → 0+ and as r → ∞.

10. Show that the boundary value problem((1 − x2)φ′)′ = −f (x), −1 < x < 1,

φ(x) bounded at x = ±1,


has as its solution

φ(x) =∫ x

0

1

1 − y2

∫ 1

yf (z)dz dy,

provided that the function f satisfies

∫ 1

−1f (z)dz = 0.

11. Suppose that the functions f (x) and φ(x) in the preceding exercise haveexpansions in terms of Legendre polynomials

f (x) =∞∑

k=0

bkPk(x), −1 < x < 1,

φ(x) =∞∑

k=0

BkPk(x), −1 < x < 1.

What is the relation between Bk and bk?

12. By applying separation of variables to the problem

∇2u = 0, 0 < ρ < a, 0 ≤ φ < π,

with u bounded at φ = 0, π and u periodic (2π) in θ , derive the follow-ing equation for the factor function �(φ):

sin(φ)(sin(φ)�′)′ − m2� + µ2 sin2(φ)� = 0,

where m = 0,1,2, . . . comes from the factor �(θ).

13. Using the change of variables x = cos(φ), �(φ) = y(x) on the equationof Exercise 12, derive a differential equation for y(x).

14. Solve the heat conduction problem

1

r

∂

∂r

(r∂u

∂r

)= 1

k

∂u

∂t, 0 < r < a, 0 < t,

∂u

∂r(a, t) = 0, 0 < t,

u(r,0) = T0 −

(r

a

)2

, 0 < r < a.


15. Solve the following potential problem in a cylinder:

1

r

∂

∂r

(r∂u

∂r

)+ ∂2u

∂z2= 0, 0 < r < a, 0 < z < b,

u(a, z) = 0, 0 < z < b,

u(r,0) = 0, u(r,b) = U0, 0 < r < a.

16. Find the solution of the heat conduction problem

1

r

∂

∂r

(r∂u

∂r

)= 1

k

∂u

∂t, 0 < r < a, 0 < t,

u(a, z) = T0, 0 < t,

u(r,0) = T1, 0 < r < a.

17. Find some frequencies of vibration of a cylinder by finding product so-lutions of the problem

1

r

∂

∂r

(r∂u

∂r

)+ ∂2u

∂z2= 1

c2

∂2u

∂t2, 0 < r < a, 0 < z < b, 0 < t,

u(r,0, t) = 0, u(r,b, t) = 0, 0 < r < a, α < t,

u(a, z, t) = 0, 0 < z < b, 0 < t.

18. Derive the given formula for the solution of the following potential equa-tion in a spherical shell:

∇2u = 0, a < ρ < b, 0 < φ < π,

u(a, φ) = f(cos(φ)

), u(b, φ) = 0, 0 < φ < π,

u(ρ,φ) =∞∑

n=0

Anb2n+1 − ρ2n+1

b2n+1 − a2n+1

(a

ρ

)n+1

Pn

(cos(φ)

),

An = 2n + 1

2

∫ 1

−1f (x)Pn(x)dx.

19. Show that the function φ(x, y) = sin(πx) sin(2πy) − sin(2πx) sin(πy)is an eigenfunction for the triangle T bounded by the lines y = 0, y = x,x = 1. That is,

∇2φ = −λ2φ in T ,

φ = 0 on the boundary of T .

What is the eigenvalue λ2 associated with φ?


20. Observe that the function φ in Exercise 19 is the difference of two differ-ent eigenfunctions of the 1×1 square (see Section 5.3) corresponding tothe same eigenvalue. Use this idea to construct other eigenfunctions forthe triangle T of Exercise 19.

21. Let T be the equilateral triangle in the xy-plane whose base is the interval0 < x < 1 of the x-axis and whose sides are segments of the lines y = √

3xand y = √

3(1 − x). Show that for n = 1,2,3, . . . , the function

φn(x, y) = sin(4nπy/

√3) + sin

(2nπ

(x − y/

√3))

− sin(2nπ

(x + y/

√3))

is a solution of the eigenvalue problem ∇2φ = −λ2φ in T , φ = 0 on theboundary of T . What are the eigenvalues λ2

n corresponding to the func-tion φn that is given? [See “The eigenvalues of an equilateral triangle,”SIAM Journal of Mathematical Analysis, 11 (1980): 819–827, by Mark A.Pinsky.]

22. In Comments and References, Section 5.11, a theorem is quoted that re-lates the least eigenvalue of a region to that of a smaller region. Confirmthe theorem by comparing the solution of Exercise 19 with the smallesteigenvalue of one-eighth of a circular disk of radius 1:

1

r

∂

∂r

(r∂φ

∂r

)+ 1

r2

∂2φ

∂θ2= −λ2φ, 0 < θ <

π

4, 0 < r < 1,

φ(r,0) = 0, φ

(r,

π

4

)= 0, 0 < r < 1,

φ(1, θ) = 0, 0 < θ <π

4.

23. Same task as Exercise 22, but use the triangle of Exercise 21 and thesmallest eigenvalue of one-sixth of a circular disk of radius 1.

24. Show that u(ρ, t) = t−3/2e−ρ2/4t is a solution of the three-dimensionalheat equation ∇2u = ∂u

∂t , in spherical coordinates.

25. For what exponent b is u(r, t) = tbe−r2/4t a solution of the two-dimensional heat equation ∇2u = ∂u

∂t ? (Use polar coordinates.)

26. Suppose that an estuary extends from x = 0 to x = a, where it meets theopen sea. If the floor of the estuary is level but its width is proportionalto x, then the water depth u(x, t) satisfies

1

x

∂

∂x

(x∂u

∂x

)= 1

gU

∂2u

∂t2, 0 < x < a, 0 < t,


where g is the acceleration of gravity and U is mean depth. The tidalmotion of the sea is represented by the boundary condition

u(a, t) = U + h cos(ωt).

Find a bounded solution of the partial differential equation that satisfiesthe boundary condition by setting

u(x, t) = U + y(x) cos(ωt).

(See Lamb, Hydrodynamics, pp. 275–276.)

27. Is there any combination of parameters for which the solution of Exer-cise 26 does not exist in the form suggested?

28. If the estuary of Exercise 26 has uniform width but variable depthh = Ux/a, then the equation for u is

∂

∂x

(x∂u

∂x

)= a

gU

∂2u

∂t2, 0 < x < a, 0 < t,

subject to the same boundary condition as in Exercise 26. Find abounded solution in the form suggested. (See Eq. (1) of Section 5.8.)

29. The equation for radially symmetric waves in n-dimensional space is

1

rn−1

∂

∂r

(rn−1 ∂u

∂r

)= 1

c2

∂2u

∂t2

where r is distance to the origin. Find product solutions of this equationthat are bounded at the origin.

30. Show that the equation of Exercise 29 has solutions of the form

u(r, t) = α(r)φ(r − ct)

for n = 1 and n = 3. [See “A simple proof that the world is three-dimensional” by Tom Morley, SIAM Review, 27 (1985): 69–71.]

31. A certain kind of chemical reactor contains particles of a solid catalystand a liquid that reacts with a gas bubbled through it. M. Chidambaran[“Catalyst mixing in bubble column slurry reactors,” Canadian Journalof Chemical Engineering, 67 (1989): 503–506] uses the following problemto model the catalyst concentration C in a cylindrical reactor:

Dr1

r

∂

∂r

(r∂C

∂r

)+ Dz

∂2C

∂z2+ U

∂C

∂z= 0, 0 < r < R, 0 < z < L,

∂C

∂r(R, z) = 0, 0 < z < L.


Here, Dr and Dz are the diffusion constants in the radial and axial di-rections, respectively. The term containing ∂C/∂z represents physicalmovement of particles at speed U .

Show that the change of variables ρ = r/R, ζ = z/L, u(ρ, ζ ) = C(r, z)leads to the equivalent equations

b1

ρ

∂

∂ρ

(ρ

∂u

∂ρ

)+ ∂2u

∂ζ 2+ p

∂u

∂ζ= 0, 0 < ρ < 1, 0 < ζ < 1,

∂u

∂ρ(1, ζ ) = 0, 0 < ζ < 1,

and identify the parameters b and p.

32. When a boundedness condition at ρ = 0 is added, product solutions ofthe foregoing equation are found to have the form u(ρ, ζ ) = R(ρ)Z(ζ ):

R0(ρ) = 1, Z0(ζ ) ={

e−pζ

1,

Rn(ρ) = J0(λnρ), Zn(ζ ) ={

em1ζ

em2ζ,

where m1 < 0 < m2 are the roots of the equation m2 +pm−λ2nb = 0 and

λn is chosen to satisfy J ′0(λn) = 0.

a. Check the details of the solution.

b. Show that the λ’s also satisfy J1(λn) = 0.

33. The solution of the problem in Exercise 31 has the form

u(ρ, ζ ) = a0e−pζ + b0 +∞∑

n=1

(anem1ζ + bnem2ζ

)J0(λnρ).

The coefficients would normally be found by applying boundary condi-tions

u(ρ,0) = f (ρ), u(ρ,1) = g(ρ), 0 < ρ < 1.

In this case, however, information is scarce. The author suggests discard-ing the solutions that do not approach 0 as ζ → ∞. The justification isthat g(ρ) is approximately 0. The solution then becomes

u(ρ, ζ ) = a0e−pζ +∞∑

n=1

anem1ζ J0(λnρ),


and the coefficients should be determined by

a0 = 2

∫ 1

0f (ρ)ρ dρ, an =

∫ 10 f (ρ)J0(λnρ)ρ dρ∫ 1

0 J20(λnρ)ρ dρ

.

The function f is known only roughly through experiment. Use thenumbers in the following table to find a0 and a1 by the trapezoidal ruleof numerical integration.

ρ 0 0.1 0.2 0.3 0.4f (ρ) 8.8 8.9 9.2 9.8 10.3J0(λ1ρ) 10 0.964 0.858 0.696 0.493

ρ 0.5 0.6 0.7 0.8 0.9 1.0f (ρ) 11.2 12.0 13.1 14.1 14.8 15J0(λ1ρ) 0.273 0.056 −0.135 −0.281 −0.373 −0.403

34. In the article “Asymptotic analysis of intraparticle diffusion in GACbatch reactors” [D.A. Lyn, Journal of Environmental Engineering, 122(1996): 1013–1022], the author analyzes chemical diffusing into a spher-ical particle, with a view to determining some parameter. The concen-tration q is modeled in dimensionless variables by

1

r2

∂

∂r

(r2 ∂q

∂r

)= ∂q

∂t, 0 < r < 1, 0 < t,

q(r, t) bounded as r → 0,

∂q

∂t(1, t) = −D

∂q

∂r(1, t).

Separate the variables and find the eigenvalue problem, assuming thatq(r, t) = R(r)T(t).

35. The eigenvalue problem that comes from Exercise 34 has a peculiarboundary condition that prevents the eigenfunctions from being orthog-onal. However, the author needs only the first terms of a series solution.Find an equation for the eigenvalues. Confirm that λ = 0 is a solution.Find the next one numerically for D = 0,1,10.

Laplace TransformC H A P T E R

6

6.1 Definition and Elementary PropertiesThe Laplace transform serves as a device for simplifying or mechanizing thesolution of ordinary and partial differential equations. It associates a functionf (t) with a function of another variable F(s) from which the original functioncan be recovered.

Let f (t) be sectionally continuous in every interval 0 ≤ t < T. The Laplacetransform of f , written L(f ) or F(s), is defined by the integral

L(f ) = F(s) =∫ ∞

0e−st f (t)dt. (1)

We use the convention that a function of t is represented by a lowercase letterand its transform by the corresponding capital letter. The variable s may bereal or complex, but in the computation of transforms by the definition, s isusually assumed to be real. Two simple examples are

L(1) =∫ ∞

0e−st · 1 dt = 1

s,

L(eat

) =∫ ∞

0e−steat dt = −e−(s−a)t

s − a

∣∣∣∣∞

0

= 1

s − a.

Not every sectionally continuous function of t has a Laplace transform, forthe defining integral may fail to converge. For instance, exp(t2) has no trans-form. However, there is a simple sufficient condition, as expressed in the fol-lowing theorem.

363

364 Chapter 6 Laplace Transform

Theorem. Let f (t) be sectionally continuous in every finite interval 0 ≤ t < T. If,for some constant k, it is true that

limt→∞ e−kt f (t) = 0,

then the Laplace transform of f exists for Re(s) > k. �

A function that satisfies the limit condition in the hypotheses of the theoremis said to be of exponential order.

The Laplace transform inherits two important properties from the integralused in its definition:

L(cf (t)

) = cL(

f (t)), c constant, (2)

L(

f (t) + g(t)) = L

(f (t)

) +L(g(t)

). (3)

By exploiting these properties, we easily determine that

L(cosh(at)

) = L[

1

2

(eat + e−at

)]

= 1

2

(1

s − a+ 1

s + a

)= s

s2 − a2,

L(sin(ωt)

) = L[

1

2i

(eiωt − e−iωt

)]

= 1

2i

(1

s − iω− 1

s + iω

)= ω

s2 + ω2.

Notice that the linearity properties work with complex constants and func-tions.

Because of the factor e−st in the definition of the Laplace transform, expo-nential multipliers are easily handled by the “shifting theorem”:

L(ebt f (t)

) =∫ ∞

0e−stebt f (t)dt

=∫ ∞

0e−(s−b)t f (t)dt = F(s − b),

where F(s) =L(f (t)). For instance, since L(sin(ωt)) = ω/(s2 + ω2),

L(ebt sin(ωt)

) = ω

(s − b)2 + ω2= ω

s2 − 2sb + b2 + ω2.

The real virtue of the Laplace transform is seen in its effect on derivatives.Suppose f (t) is continuous and has a sectionally continuous derivative f ′(t).

Chapter 6 Laplace Transform 365

Then by definition

L(

f ′(t)) =

∫ ∞

0e−st f ′(t)dt.

Integrating by parts, we get

L(

f ′(t)) = e−st f (t)

∣∣∞0

−∫ ∞

0(−s)e−st f (t)dt.

If f (t) is of exponential order, e−st f (t) must tend to 0 as t tends to infinity (forlarge enough s), so the foregoing equation becomes

L(

f ′(t)) = −f (0) + s

∫ ∞

0e−st f (t)dt

= −f (0) + sL(

f (t)).

(If f (t) has a jump at t = 0, f (0) is to be interpreted as f (0+).)Similarly, if f and f ′ are continuous, f ′′ is sectionally continuous; and if all

three functions are exponential order, then

L(

f ′′(t)) = −f (0) + sL

(f ′(t)

)= −f (0) − sf (0) + s2L

(f (t)

).

An easy generalization extends this formula to the nth derivative,

L[

f (n)(t)] = −f (n−1)(0) − sf (n−2)(0) − · · · − sn−1f (0) + snL

(f (t)

), (4)

on the assumption that f and its first n − 1 derivatives are continuous, f (n) issectionally continuous, and all are of exponential order.

We may apply Eq. (4) to the function f (t) = tk, k being a nonnegative inte-ger. Here we have

f (0) = f ′(0) = · · · = f (k−1)(0) = 0, f (k)(0) = k!, f (k+1)(t) = 0.

Thus, Eq. (4) with n = k + 1 yields

0 = −k! + sk+1L(tk

),

or

L(tk

) = k!

sk+1.

A different application of the derivative rule is used to transform integrals. Iff (t) is sectionally continuous, then

∫ t0 f (t′)dt′ is a continuous function, equal

to zero at t = 0, and has derivative f (t). Hence

L(

f (t)) = sL

[∫ t

0f (t′)dt′

],


L(f ) = F(s) =∫ ∞

0

e−st f (t)dt

L(cf (t)) = cL( f (t))

L( f (t) + g(t)) = L( f (t)) +L( g(t))

L( f ′(t)) = −f (0) + sF(s)

L( f ′′(t)) = −f ′(0) − sf (0) + s2F(s)

L( f (n)(t)) = −f (n−1)(0) − sf (n−2)(0) − · · · − sn−1f (0) + snF(s)

L(ebt f (t)) = F(s − b)

L(∫ t

0

f (t′)dt′)

= 1

sF(s) L

(1

tf (t)

)= ∫ ∞

s F(s′)ds′

L(tf (t)) = −dF

ds

Table 1 Properties of the Laplace transform

or

L[∫ t

0f (t′)dt′

]= 1

sL

(f (t)

). (5)

Differentiation and integration with respect to s may produce transforma-tions of previously inaccessible functions. We need the two formulas

−de−st

ds= te−st,

∫ ∞

se−s′t ds′ = 1

te−st

to derive the results

L(tf (t)

) = −dF(s)

ds, L

(1

tf (t)

)=

∫ ∞

sF(s′)ds′. (6)

(Note that, unless f (0) = 0, the transform of f (t)/t will not exist.) Examplesof the use of these formulas are

L(t sin(ωt)

) = − d

ds

(ω

s2 + ω2

)= 2sω

(s2 + ω2)2,

L(

sin(t)

t

)=

∫ ∞

s

ds′

s′2 + 1= π

2− tan−1(s) = tan−1

(1

s

).

Significant properties of the Laplace transform are summarized in Table 1.When a problem is solved by use of Laplace transforms, a prime difficulty

is computation of the corresponding function of t. Methods for computingthe “inverse transform” f (t) = L−1(F(s)) include integration in the complexplane, convolution, partial fractions (discussed in Section 6.2), and tables of

Chapter 6 Laplace Transform 367

f (t) F(s) f (t) F(s)

0 0 tk k!

sk+1

11

sebt cos(ωt)

s − b

s2 − 2bs + b2 + ω2

eat 1

s − aebt sin(ωt)

ω

s2 − 2bs + b2 + ω2

cosh(at)s

s2 − a2ebt tk k!

(s − b)k+1

sinh(at)a

s2 − a2eat − 1

a

s(s − a)

cos(ωt)s

s2 + ω2t cos(ωt)

s2 − ω2

(s2 + ω2)2

sin(ωt)ω

s2 + ω2t sin(ωt)

2sω

(s2 + ω2)2

t1

s2

Table 2 Laplace transforms

transforms. The last method, which involves the least work, is the most popu-lar. The transforms in Table 2 were all calculated from the definition or by useof formulas in this section.

E X E R C I S E S

1. By using linearity and the transform of eat , compute the transform of eachof the following functions.

a. sinh(at);

d. sin(ωt − φ);

b. cos(ωt);

e. e2(t+1);

c. cos2(ωt);

f. sin2(ωt).

2. Use differentiation with respect to t to find the transform of

a. teat from L(eat),

b. sin(ωt) from L(cos(ωt)),

c. cosh(at) from L(sinh(at)).

3. Compute the transform of each of the following directly from the defini-tion.

a. f (t) ={0, 0 < t < a,

1, a < t;


b. f (t) ={0, 0 < t < a,

1, a < t < b,0, b < t;

c. f (t) ={ t, 0 < t < a,

a, a < t.

4. The Heaviside step function is defined by the formula

Ha(t) ={1, t > a,

0, t < a.

Assuming a ≥ 0, show that the Laplace transform of Ha is

L(Ha(t)

) = e−as

s.

5. Use completion of square and the shifting theorem to find the inverse trans-form of

a.1

s2 + 2s, b.

s + 1

s2 + 2s + 2, c.

1

s2 + 2as + b2, b > a.

6. Find the Laplace transform of the square-wave function

f (t) ={1, 0 < x < a,

0, a < x < 2a,f (x + 2a) = f (x).

Hint: Break up the integral as shown in the following, evaluate the integrals,and add up a geometric series:

F(s) =∞∑

n=0

∫ 2(n+1)a

2naf (t)e−st dt.

7. Use any method to find the inverse transform of the following.

a.1

(s − a)(s − b);

c.s2

(s2 + ω2)2;

b.s

(s2 − a2)2;

d.1

(s − a)3; e.

1 − e−s

s.

8. Use any theorem or formula to find the transform of the following.

a.1 − cos(ωt)

t;

c. t2e−at ;

b.∫ t

0

sin(at′)t′ dt′;

d. t cos(ωt); e. sinh(at) sin(ωt).

9. Find the inverse transform of these functions of s by any method.

6.2 Partial Fractions and Convolutions 369

a.1

(s2 + ω2)2;

c.s2

(s2 + ω2)2;

b.s

(s2 + ω2)2;

d.s3

(s2 + ω2)2.

6.2 Partial Fractions and Convolutions

Because of the formula for the transform of derivatives, the Laplace transformfinds important application to linear differential equations with constant co-efficients, subject to initial conditions. In order to solve the simple problem

u′ + au = 0, u(0) = 1,

we transform the entire equation, obtaining

L(u′) + aL(u) = 0

or

sU − 1 + aU = 0,

where U = L(u). The derivative has been “transformed out,” and U is deter-mined by simple algebra to be

U(s) = 1

s + a.

By consulting Table 2 we find that u(t) = e−at .Equations of higher order can be solved in the same way. When trans-

formed, the problem

u′′ + ω2u = 0, u(0) = 1, u′(0) = 0

becomes

s2U − s · 1 − 0 + ω2 U = 0.

Note how both initial conditions have been incorporated into this one equa-tion. Now we solve the transformed equation algebraically to find

U(s) = s

s2 + ω2,

the transform of cos(ωt) = u(t).


In general we may outline our procedure as follows:

Originalproblem

L

Solution oforiginal problem

Transformedproblem

Solution oftransformed

problem

L−1

In the step marked L−1, we must compute the function of t to which the solu-tion of the transformed problem corresponds. This is the difficult part of theprocess. The key property of the inverse transform is its linearity, as expressedby

L−1(c1F1(s) + c2F2(s)

) = c1L−1(F1(s)

) + c2L−1(F2(s)

).

This property allows us to break down a complicated transform into a sum ofsimple ones.

A simple mass–spring–damper system leads to the initial value problem

u′′ + au′ + ω2u = 0, u(0) = u0, u′(0) = u1,

whose transform is

s2U − su0 − u1 + a(sU − u0) + ω2U = 0.

Determination of U gives it as the ratio of two polynomials:

U(s) = su0 + (u1 + au0)

s2 + as + ω2.

Although this expression is not in Table 2, it can be worked around to a func-tion of s + a/2, whose inverse transform is available. The shift theorem thengives u(t). There is, however, a better way.

The inversion of a rational function of s (that is, the ratio of two polynomi-als) can be accomplished by the technique of “partial fractions.” Suppose wewish to compute the inverse transform of

U(s) = cs + d

s2 + as + b.

The denominator has two roots, r1 and r2, which we assume for the momentto be distinct. Thus

s2 + as + b = (s − r1)(s − r2),

U(s) = cs + d

(s − r1)(s − r2).


The rules of elementary algebra suggest that U can be written as a sum,

cs + d

(s − r1)(s − r2)= A1

s − r1+ A2

s − r2, (1)

for some choice of A1 and A2. Indeed, by finding the common denominatorform for the right-hand side and matching powers of s in the numerator, weobtain

cs + d

(s − r1)(s − r2)= A1(s − r2) + A2(s − r1)

(s − r1)(s − r2),

c = A1 + A2, d = −A1r2 − A2r1.

When A1 and A2 are determined, the inverse transform of the right-hand sideof Eq. (1) is easily found:

L−1

(A1

s − r1+ A2

s − r2

)= A1 exp(r1t) + A2 exp(r2t).

For a specific example, suppose that

U(s) = s + 4

s2 + 3s + 2.

The roots of the denominator are r1 = −1 and r2 = −2. Thus

s + 4

s2 + 3s + 2= A1

s + 1+ A2

s + 2= (A1 + A2)s + (2A1 + A2)

(s + 1)(s + 2).

We find A1 = 3, A2 = −2. Hence

L−1

(s + 4

s2 + 3s + 2

)=L−1

(3

s + 1− 2

s + 2

)= 3e−t − 2e−2t .

A little calculus takes us much further. Suppose that U has the form

U(s) = q(s)

p(s),

where p and q are polynomials and the degree of q is less than the degree of p.Assume that p has distinct roots r1, . . . , rk:

p(s) = (s − r1)(s − r2) · · · (s − rk).

We try to write U in the fraction form

U(s) = A1

s − r1+ A2

s − r2+ · · · + Ak

s − rk= q(s)

p(s).


The algebraic determination of the A’s is very tedious, but notice that

(s − r1)q(s)

p(s)= A1 + A2

s − r1

s − r2+ · · · + Ak

s − r1

s − rk.

If s is set equal to r1, the right-hand side is just A1. The left-hand side becomes0/0, but L’Hôpital’s rule gives

lims→r1

(s − r1)q(s)

p(s)= lim

s→r1

(s − r1)q′(s) + q(s)

p′(s)= q(r1)

p′(r1).

Therefore A1 and all the other A’s are given by

Ai = q(ri)

p′(ri).

Consequently, our rational function takes the form

q(s)

p(s)= q(r1)

p′(r1)

1

s − r1+ · · · + q(rk)

p′(rk)

1

s − rk.

From this point we can easily obtain the inverse transform, as expressed in theconclusion of the following theorem.

Theorem 1. Let p and q be polynomials, q of lower degree than p, and let p haveonly simple roots, r1, r2, . . . , rk. Then

L−1

(q(s)

p(s)

)= q(r1)

p′(r1)exp(r1t) + · · · + q(rk)

p′(rk)exp(rkt). (2)

(Equation (2) is known as Heaviside’s formula.) �

Let us apply the theorem to the example in which q(s) = s + 4, p(s) = s2 +3s + 2, p′(s) = 2s + 3. Then

L−1

(s + 4

s2 + 3s + 2

)= −1 + 4

2(−1) + 3e−t + −2 + 4

2(−2) + 3e−2t = 3e−t − 2e−2t .

In nonhomogeneous differential equations also, the Laplace transform is auseful tool. To solve the problem

u′ + au = f (t), u(0) = u0,

we again transform the entire equation, obtaining

sU − u0 + aU = F(s),

U(s) = u0

s + a+ 1

s + aF(s).


The first term in this expression is recognized as the transform of u0e−at . IfF(s) is a rational function, partial fractions may be used to invert the secondterm. However, we can identify that term by solving the problem another way.For example,

eat(u′ + au) = eat f (t),(ueat

)′ = eat f (t),

ueat =∫ t

0eat′ f

(t′)dt′ + c,

u(t) =∫ t

0e−a(t−t′)f

(t′)dt′ + ce−at .

The initial condition requires that c = u0. On comparing the two results, wesee that

L[∫ t

0e−a(t−t′)f

(t′)dt′

]= 1

s + aF(s).

Thus the transform of the combination of e−at and f (t) on the left is the prod-uct of the transforms of e−at and f (t). This simple result can be generalized inthe following way.

Theorem 2. If g(t) and f (t) have Laplace transforms G(s) and F(s), respectively,then

L[∫ t

0g(t − t′)f

(t′)dt′

]= G(s)F(s). (3)

(This is known as the convolution theorem.) �

The integral on the left is called the convolution of g and f , written

g(t) ∗ f (t) =∫ t

0g(t − t′)f

(t′)dt′.

It can be shown that the convolution follows these rules:

g ∗ f = f ∗ g, (4a)

f ∗ ( g ∗ h) = ( f ∗ g) ∗ h, (4b)

f ∗ ( g + h) = f ∗ g + f ∗ h. (4c)

The convolution theorem provides an important device for invertingLaplace transforms, which we shall apply to find the general solution of thenonhomogeneous problem

u′′ − au = f (t), u(0) = u0, u′(0) = u1.


The transformed equation is readily solved, yielding

U(s) = su0 + u1

s2 − a+ 1

s2 − aF(s).

Because 1/(s2 −a) is the transform of sinh(√

at)/√

a, we easily determine thatu is

u(t) = u0 cosh(√

at) + u1√

asinh

(√at

) +∫ t

0

sinh(√

a(t − t′))√a

f(t′)dt′. (5)

A slightly different problem occurs if the mass in a spring–mass system isstruck while the system is in motion. The mathematical model of the systemmight be

u′′ + ω2u = f (t), u(0) = u0, u′(0) = u1,

where f (t) = F0 for t0 < t < t1 and f (t) = 0 for other values. The transform ofu is

U(s) = su0 + u1

s2 + ω2+ 1

s2 + ω2F(s).

The inverse transform of U(s) is then

u(t) = u0 cos(ωt) + u1sin(ωt)

ω+

∫ t

0

sin(ω(t − t′))ω

f(t′)dt′.

The convolution in this case is easy to calculate:

∫ t

0

sin(ω(t − t′))ω

f(t′)dt′

=

0, t < t0,

F01 − cos(ω(t − t0))

ω2, t0 < t < t1,

F0cos(ω(t − t1)) − cos(ω(t − t0))

ω2, t1 < t.

E X E R C I S E S

1. Solve these initial value problems.

a. u′ − 2u = 0, u(0) = 1;

b. u′ + 2u = 0, u(0) = 1;

c. u′′ + 4u′ + 3u = 0, u(0) = 1, u′(0) = 0;

d. u′′ + 9u = 0, u(0) = 0, u′(0) = 1.


2. Solve the initial value problem

u′′ + 2au′ + u = 0, u(0) = u0, u′(0) = u1

in these three cases: 0 < a < 1, a = 1, a > 1.

3. Solve these nonhomogeneous problems with zero initial conditions.

a. u′ + au = 1;

c. u′′ + 4u = sin(t);

e. u′′ + 2u′ = 1 − e−t ;

b. u′′ + u = t;

d. u′′ + 4u = sin(2t);

f. u′′ − u = 1.

4. Complete the square in the denominator and use the shift theorem[F(s − a) =L(eat f (t))] to invert

U(s) = su0 + (u1 + 2au0)

s2 + 2as + ω2.

There are three cases, corresponding to

ω2 − a2 > 0, = 0, < 0.

5. Use partial fractions to invert the following transforms.

a.1

s2 − 4;

c.(s + 3)

s(s2 + 2);

b.1

s2 + 4;

d.4

s(s + 1).

6. Prove properties (4a) and (4c) of the convolution.

7. Compute the convolution f ∗ g for

a. f (t) = 1, g(t) = sin(t);

b. f (t) = et , g(t) = cos(ωt);

c. f (t) = t, g(t) = sin(t).

8. Demonstrate the following properties of convolution either directly or byusing Laplace transform.

a. 1 ∗ f ′(t) = f (t) − f (0);

b. (t ∗ f (t))′′ = f (t);

c. ( f ∗ g)′ = f ′ ∗ g = f ∗ g ′, if f (0) = g(0) = 0.


6.3 Partial Differential Equations

In applying the Laplace transform to partial differential equations, we treatvariables other than t as parameters. Thus, the transform of a function u(x, t)is defined by

L(u(x, t)

) =∫ ∞

0e−stu(x, t)dt = U(x, s).

For instance, we easily find the transforms

L(e−at sin(πx)

) = 1

s + asin(πx),

L(sin(x + t)

) = s sin(x) + cos(x)

s2 + 1.

The transform U naturally is a function not only of s but also of the “untrans-formed” variable x. We assume that derivatives or integrals with respect to theuntransformed variable pass through the transform

L(

∂u

∂x

)=

∫ ∞

0

∂u(x, t)

∂xe−st dt

= ∂

∂x

∫ ∞

0u(x, t)e−st dt = ∂

∂x

(U(x, s)

).

If we wish to focus on the role of x as a variable and keep s in the backgroundas a parameter, we might use the symbol for the ordinary derivative:

L(

∂u

∂x

)= dU

dx.

The rule for transforming a derivative with respect to t can be found, asbefore, with integration by parts:

L(

∂u

∂t

)= sL

(u(x, t)

) − u(x,0).

If the Laplace transform is applied to a boundary value–initial value prob-lem in x and t, all time derivatives disappear, leaving an ordinary differentialequation in x. We shall illustrate this technique with some trivial examples.Incidentally, we assume from here on that problems have been prepared (forexample, by dimensional analysis) so as to eliminate as many parameters aspossible.

6.3 Partial Differential Equations 377

Example 1.

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 1, u(1, t) = 1, 0 < t,

u(x,0) = 1 + sin(πx), 0 < x < 1.

The partial differential equation and the boundary conditions (that is, every-thing that is valid for t > 0) are transformed, while the initial condition isincorporated by the transform

d2U

dx2= sU − (

1 + sin(πx)), 0 < x < 1,

U(0, s) = 1

s, U(1, s) = 1

s.

This boundary value problem is solved to obtain

U(x, s) = 1

s+ sin(πx)

s + π2.

We direct our attention now to U as a function of s. Because sin(πx) is a con-stant with respect to s, tables may be used to find

u(x, t) = 1 + sin(πx) exp(−π2t

). �

Example 2.

∂2u

∂x2= ∂2u

∂t2, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = sin(πx), 0 < x < 1,

∂u

∂t(x,0) = − sin(πx), 0 < x < 1.

Under transformation the problem becomes

d2U

dx2= s2U − s sin(πx) + sin(πx), 0 < x < 1,

U(0, s) = 0, U(1, s) = 0.

The function U is found to be

U(x, s) = s − 1

s2 + π2sin(πx),


from which we find the solution,

u(x, t) =(

cos(π t) − 1

πsin(π t)

)sin(πx). �

Example 3.Now we consider a problem that we know to have a more complicated solu-tion:

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 1, u(1, t) = 1, 0 < t,

u(x,0) = 0, 0 < x < 1.

The transformed problem is

d2U

dx2= sU, 0 < x < 1,

U(0, s) = 1

s, U(1, s) = 1

s.

The general solution of the differential equation is well known to be a combi-nation of sinh(

√sx) and cosh(

√sx). Application of the boundary conditions

yields

U(x, s) = 1

scosh

(√sx

) + (1 − cosh(√

s)) sinh(√

sx)

s sinh(√

s)

= sinh(√

sx) + sinh(√

s(1 − x))

s sinh(√

s).

This function rarely appears in a table of transforms. However, by extendingthe Heaviside formula, we can compute an inverse transform.

When U is the ratio of two transcendental functions (not polynomials) of s,we wish to write

U(x, s) =∑

An(x)1

s − rn.

In this formula, the numbers rn are values of s for which the “denominator” ofU is zero, or, rather, for which |U(x, s)| becomes infinite; the An are functionsof x but not s. From this form we expect to determine

u(x, t) =∑

An(x) exp(rnt).

This solution should be checked for convergence.The hyperbolic sine (also the cosh, cos, sin, and exponential functions) is

not infinite for any finite value of its argument. Thus U(x, s) becomes infinite


only where s or sinh(√

s) is zero. Because sinh(√

s) = 0 has no real root besideszero, we seek complex roots by setting

√s = ξ + iη (ξ and η real).

The addition rules for hyperbolic and trigonometric functions remain validfor complex arguments. Furthermore, we know that

cosh(iA) = cos(A), sinh(iA) = i sin(A).

By combining the addition rule and these identities we find

sinh(ξ + iη) = sinh(ξ) cos(η) + i cosh(ξ) sin(η).

This function is zero only if both the real and imaginary parts are zero. Thus ξ

and η must be chosen to satisfy simultaneously

sinh(ξ) cos(η) = 0, cosh(ξ) sin(η) = 0.

Of the four possible combinations, only sinh(ξ) = 0 and sin(η) = 0 producesolutions. Therefore ξ = 0 and η = ±nπ (n = 0,1,2, . . .); whence

√s = ±inπ, s = −n2π2.

Recall that only the value of s, not the value of√

s, is significant.Finally then, we have located r0 = 0, and rn = −n2π2 (n = 1,2, . . .). We

proceed to find the An(x) by the same method used in Section 6.2. The com-putations are done piecemeal and then the solution is assembled.

Part a. (r0 = 0.) In order to find A0 we multiply both sides of our proposedpartial fractions development

U(x, s) =∞∑

n=0

An(x)1

s − rn

by s − r0 = s and take the limit as s approaches r0 = 0. The right-hand sidegoes to A0. On the left-hand side we have

lims→0

ssinh(

√sx) + sinh(

√s(1 − x))

s sinh(√

s)= x + 1 − x = 1 = A0(x).

Thus the part of u(x, t) corresponding to s = 0 is 1 · e0t = 1, which is easilyrecognized as the steady-state solution.

Part b. (rn = −n2π2, n = 1,2, . . . .) For these cases, we find

An = q(rn)

p′(rn),


where q and p are the obvious choices. We take√

rn = +inπ in all calculations:

p′(s) = sinh(√

s) + s

1

2√

scosh

(√s),

p′(rn) = 1

2inπ cosh(inπ) = 1

2inπ cos(nπ),

q(rn) = sinh(inπx) + sinh(inπ(1 − x)

)= i

[sin(nπx) + sin

(nπ(1 − x)

)].

Hence the portion of u(x, t) that arises from each rn is

An(x) exp(rnt) = 2sin(nπx) + sin(nπ(1 − x))

nπ cos(nπ)exp

(−n2π2t).

Part c. On assembling the various pieces of the solution, we get

u(x, t) = 1 + 2

π

∞∑1

sin(nπx) + sin(nπ(1 − x))

n cos(nπ)exp

(−n2π2t).

The same solution would be found by separation of variables but in a slightlydifferent form. �

Example 4.Now consider the wave problem

∂2u

∂x2= ∂2u

∂t2, 0 < x < 1, 0 < t,

u(0, t) = 0,∂u(1, t)

∂x= 0, 0 < t,

u(x,0) = 0,∂u(x,0)

∂t= x, 0 < x < 1.

The transformed problem is

d2U

dx2= s2U − x, 0 < x < 1,

U(0, s) = 0, U ′(1, s) = 0,

and its solution (by undetermined coefficients or otherwise) gives

U(x, s) = sx cosh(s) − sinh(sx)

s3 cosh(s).

The numerator of this function is never infinite. The denominator is zero ats = 0 and s = ±i(2n − 1)π/2 (n = 1,2, . . .). We shall again use the Heavisideformula to determine the inverse transform of U .


Part a. (r0 = 0.) The limit as s approaches zero of sU(x, s) may be found byL’Hôpital’s rule or by using the Taylor series for sinh and cosh. From the latter,

sU(x, s) = sx(1 + s2

2 + · · ·) − (sx + s3x3

6 + · · ·)s2

(1 + s2

2 + · · ·)= s3

(x2 − x3

6 − · · ·)s2

(1 + s2

2 + · · ·) → 0.

Thus, in spite of the formidable appearance of s3 in the denominator, s = 0 isnot really a significant value and contributes nothing to u(x, t).

Part b. It is convenient to take the remaining roots in pairs. We label

±i(2n − 1)π

2= ±iρn.

The derivative of the denominator is

p′(s) = 3s2 cosh(s) + s3 sinh(s),

p′(±iρn) = ±i3ρ3n sinh(±iρn)

= ρ3n sin(ρn)

since sinh(iρ) = i sin(ρ) and (±i)4 = 1. The contribution of these two rootstogether may be calculated using the exponential definition of sine:

q(iρn)

p′(iρn)exp(iρnt) + q(−iρn)

r′(−iρn)exp(−iρnt)

= − sinh(iρnx) exp(iρnt) + sinh(iρnx) exp(−iρnt)

ρ3n sin(ρn)

= sin(ρnx)

ρ3n sin(ρn)

i(− exp(iρnt) + exp(−iρnt)

)

= 2 sin(ρnx) sin(ρnt)

ρ3n sin(ρn)

.

Part c. The final form of u(x, t), found by adding up all the contributionsfrom Part b, is the same as would be found by separation of variables

u(x, t) = 2∞∑1

sin(ρnx) sin(ρnt)

ρ3n sin(ρn)

. �


E X E R C I S E S

1. Find all values of s, real and complex, for which the following functions arezero.

a. cosh(√

s);

c. sinh(s);

e. cosh(s) + s sinh(s).

b. cosh(s);

d. cosh(s) − s sinh(s);

2. Find the inverse transforms of the following functions in terms of an infi-nite series.

a.1

stanh(s); b.

sinh(sx)

s cosh(s).

3. Find the transform U(x, s) of the solution of each of the following prob-lems.

a.∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = t, 0 < t,

u(x,0) = 0, 0 < x < 1;

b.∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = e−t, 0 < t,

u(x,0) = 1, 0 < x < 1.

4. Solve each of the problems in Exercise 3, inverting the transform by meansof the extended Heaviside formula.

5. Solve each of the following problems by Laplace transform methods.

a.∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 1, 0 < t,

u(x,0) = 0, 0 < x < 1;

b.∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = 1, 0 < x < 1.

6.4 More Difficult Examples 383

6.4 More Difficult ExamplesThe technique of separation of variables, once mastered, seems more straight-forward than the Laplace transform. However, when time-dependent bound-ary conditions or inhomogeneities are present, the Laplace transform offersa distinct advantage. Following are some examples that display the power oftransform methods.

Example 1.A uniform insulated rod is attached at one end to an insulated container offluid. The fluid is circulated so well that its temperature is uniform and equalto that at the end of the rod. The other end of the rod is maintained at a con-stant temperature. A dimensionless initial value–boundary value problem thatdescribes the temperature in the rod is

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

∂u(0, t)

∂x= γ

∂u(0, t)

∂t, u(1, t) = 1, 0 < t,

u(x,0) = 0, 0 < x < 1.

The transformed problem and its solution are

d2U

dx2= sU, 0 < x < 1,

dU

dx(0, s) = sγ U(0, s), U(1, s) = 1

s,

U(x, s) = cosh(√

sx) + √sγ sinh(

√sx)

s(cosh(√

s) + √sγ sinh(

√s))

= q(s)

p(s).

Aside from s = 0, the denominator has no real zeros. Thus we again searchfor complex zeros by employing

√s = ξ + iη. The real and imaginary parts of

the denominator are to be computed by using the addition formulas for coshand sinh. The requirement that both real and imaginary parts be zero leads tothe equations(

cosh(ξ) + ξγ sinh(ξ))

cos(η) − ηγ cosh(ξ) sin(η) = 0, (1)

ηγ sinh(ξ) cos(η) + (sinh(ξ) + ξγ cosh(ξ)

)sin(η) = 0. (2)

We may think of these as simultaneous equations in sin(η) and cos(η). Becausesin2(η) + cos2(η) = 1, the system has a solution only when its determinant iszero. Thus after some algebra, we arrive at the condition(

1 + ξ 2γ 2 + η2γ 2)

sinh(ξ) cosh(ξ) + ξγ(sinh2

(ξ) + cosh2(ξ)

) = 0.


The only solution occurs when ξ = 0, for otherwise both terms of this equa-tion have the same sign.

After setting ξ = 0, we find that Eq. (1) reduces to

tan(η) = 1

ηγ,

for which there is an infinite number of solutions. We shall number the posi-tive solutions η1, η2, . . . . Now, we have found that the roots of p(s) are r0 = 0and rk = (iηk)

2 = −η2k . The computation of the inverse transform follows.

Part a. (r0 = 0.) The limit of sU(x, s) as s tends to zero is easily found to be1. Thus this root contributes 1 · e0t = 1 to u(x, t).

Part b. (rk = −η2k .) First, we compute

p′(s) = cosh(√

s) + √

sγ sinh(√

s) + 1

2

√s(1 + γ ) sinh

(√s) + 1

2γ s cosh

(√s).

Using the fact that cosh(√

rk) + √rkγ sinh(

√rk) = 0, we may reduce the fore-

going to

p′(rk) = −1

2γ

(1 + γ + η2

kγ2)

cos(ηk).

Hence the contribution to u(x, t) of rk is

q(rk)

p′(rk)exp(rkt) = −2γ

cos(ηkx) − ηkγ sin(ηkx)

(1 + γ + η2kγ

2) cos(ηk)exp

(−η2kt

).

Part c. The construction of the final solution is left to the reader. We notethat an attempt to solve this problem by separation of variables would finddifficulties, for the eigenfunctions are not orthogonal. �

Example 2.Sometimes one is interested not in the complete solution of a problem, butonly in part of it. For example, in the problem of heat conduction in a semi-infinite solid with time-varying boundary conditions, we may seek that part ofthe solution that persists after a long time. (This may or may not be a steady-state solution.) Any initial condition that is bounded in x gives rise only totransient temperatures; these being of no interest, we assume a zero initial con-dition. Thus, the problem to be studied is

∂2u

∂x2= ∂u

∂t, 0 < x < ∞, 0 < t,

u(0, t) = f (t), 0 < t,

u(x,0) = 0, 0 < x < ∞.


The transformed equation and its general solution are

d2U

dx2= sU, 0 < x, U(0, s) = F(s),

U(x, s) = A exp(−√

sx) + B exp

(√sx

).

We make two further assumptions about the solution: first, that u(x, t) isbounded as x tends to infinity and, second, that

√s means the square root

of s that has a nonnegative real part. Under these two assumptions, we mustchoose B = 0 and A = F(s), making

U(x, s) = F(s) exp(−√

sx).

In order to find the persistent part of u(x, t), we apply the Heaviside inver-sion formula to those values of s having nonnegative real parts, because a valueof s with negative real part corresponds to a function containing a decayingexponential — a transient. To understand this fact, consider the pair

f (t) = 1 − e−βt + α sin(ωt),

F(s) = 1

s− 1

s + β+ αω

s2 + ω2.

Now we return to the original problem with this choice for f (t). The valuesof s for which U(x, s) = F(s) exp(−√

sx) becomes infinite are 0, ±iω, −β . Thelast value is discarded, because it is negative. Thus, the persistent part of thesolution is given by

A0e0t + A1eiωt + A2e−iωt,

and the coefficients are found from

A0 = lims→0

[sF(s) exp

(−√sx

)] = 1,

A1 = lims→iω

[(s − iω)F(s) exp

(−√sx

)] = α

2iexp

(−√iωx

),

A2 = lims→−iω

[(s + iω)F(s) exp

(−√sx

)] = − α

2iexp

(−√−iωx).

We also need to know that the roots of ±i with positive real part are

√i = 1√

2(1 + i),

√−i = 1√2(1 − i).

Thus the function we seek is


1 + 1

2iexp

[iωt −

√ω

2(1 + i)x

]− α

2iexp

[−iωt −

√ω

2(1 − i)x

]

= 1 + α exp

(−

√ω

2x

)sin

(ωt −

√ω

2x

). �

Example 3.If a steel wire is exposed to a sinusoidal magnetic field, the boundary value–initial value problem that describes its displacement is

∂2u

∂x2= ∂2u

∂t2− sin(ωt), 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 0, 0 < x < 1.

The nonhomogeneity in the partial differential equation represents the effectof the force due to the field. The transformed equation and its solution are

d2U

dx2= s2U − ω

s2 + ω2, 0 < x < 1,

U(0, s) = 0, U(1, s) = 0,

U(x, s) = ω

s2(s2 + ω2)

cosh( 12 s) − cosh

(s( 1

2 − x))

cosh( 12 s)

.

Several methods are available for the inverse transformation of U . An obvi-ous one would be to compute

v(x, t) =L−1

(cosh( 1

2 s) − cosh(s( 1

2 − x))

s2 cosh( 12 s)

)

and write u(x, t) as a convolution

u(x, t) =∫ t

0sin

(ω

(t − t′))v(

x, t′)dt′.

The details of this development are left as an exercise.We could also use the Heaviside formula. The application is now rou-

tine, except in the interesting case where cosh(iω/2) = 0, that is, where ω =(2n − 1)π , one of the natural frequencies of the wire.

Let us suppose ω = π , so

U(x, s) = π

s2(s2 + π2)


(s( 1

2 − x))

cosh( 12 s)

.


At the points s = 0, s = ±iπ , s = ±(2n − 1)iπ , n = 2,3, . . . , U(x, s) becomesundefined. The computation of the parts of the inverse transform correspond-ing to the points other than ±iπ is easily carried out. However, at these twotroublesome points, our usual procedure will not work. Instead of expecting apartial-fraction decomposition containing

A−1

s + iπ+ A1

s − iπ

and other terms of the same sort, we must seek terms like

A−1(s + iπ) + B−1

(s + iπ)2+ A1(s − iπ) + B1

(s − iπ)2,

whose contribution to the inverse transform of U would be

A−1e−iπ t + B−1te−iπ t + A1eiπ t + B1teiπ t .

One can compute A1 and B1, for example, by noting that

B1 = lims→iπ

[(s − iπ)2U(x, s)

],

A1 = lims→iπ

{(s − iπ)

[U(x, s) − B1

(s − iπ)2

]}

and similarly for A−1 and B−1. The limit for B1 is not too difficult. For example,

B1 = lims→iπ

{π

s2(s + iπ)


(s( 1

2 − x))

(cosh( 1

2 s))/(s − iπ)

}

= π

−π2(2iπ)

cosh( 12 iπ) − cosh

(iπ( 1

2 − x))

12 sinh( 1

2 iπ)

= −1

π2cos

(π

(1

2− x

))= −1

π2sin(πx).

The limit for A1 is rather more complicated but may be computed byL’Hôpital’s rule. Nevertheless, because B−1 = B1, we already see that u(x, t)contains the term

B1teiπ t + B−1te−iπ t = − 2t

π2sin(πx) cos(π t),

whose amplitude increases with time. This, of course, is the expected reso-nance phenomenon. �


E X E R C I S E S

1. Find the persistent part of the solution of the heat problem

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

∂u

∂x(0, t) = 0,

∂u

∂x(1, t) = 1, 0 < t,

u(x,0) = 0, 0 < x < 1.

2. Verify that the persistent part of the solution to Example 2 actually satisfiesthe heat equation. What boundary condition does it satisfy?

3. Find the function v(x, t) whose transform is


(s( 1

2 − x))

s2 cosh( 12 s)

.

What boundary value–initial value problem does v(x, t) satisfy?

4. Solve

∂2u

∂x2= ∂2u

∂t2, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 1, 0 < x < 1.

5. a. Solve for ω �= π :

∂2u

∂x2= ∂2u

∂t2− sin(πx) sin(ωt), 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 0, 0 < x < 1.

b. Examine the special case ω = π .

6. Obtain the complete solution of Example 1 and verify that it satisfies theboundary conditions and the heat equation.


7. a. Solve

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 1 − e−at, 0 < t,

u(x,0) = 0, 0 < x < 1.

b. Examine the special case where a = n2π2 for some integer n.


The real development of the Laplace transform began in the late nineteenthcentury, when engineer Oliver Heaviside invented a powerful, but unjustified,symbolic method for studying the ordinary and partial differential equationsof mathematical physics. By the 1920s, Heaviside’s method had been legitima-tized and recast as the Laplace transform that we now use. Later generalizationsare Schwartz’s theory of distributions (1940s) and Mikusinski’s operationalcalculus (1950s). The former seems to be the more general. Both theories givean interpretation of F(s) = 1, which is not the Laplace transform of any func-tion, in the sense we use.

There are a number of other transforms, under the names of Fourier, Mellin,Hänkel, and others, similar in intent to the Laplace transform, in which someother function replaces e−st in the defining integral. Operational Mathematics,by Churchill, has more information about the applications of transforms. Ex-tensive tables of transforms will be found in Tables of Integral Transforms byErdelyi et al. (See the Bibliography.)

Miscellaneous Exercises

1. Solve the heat conduction problem

∂2u

∂x2− γ 2(u − T) = ∂u

∂t, 0 < x < 1, 0 < t,

∂u

∂x(0, t) = 0,

∂u

∂x(1, t) = 0, 0 < t,

u(x,0) = T0, 0 < x < 1.


2. Find the “persistent part” of the solution of

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

∂u

∂x(0, t) = 0, u(1, t) = t, 0 < t,

u(x,0) = 0, 0 < x < 1.

3. Find the complete solution of the problem in Exercise 2.

4. A solid object and a surrounding fluid exchange heat by convection. Thetemperatures u1 and u2 are governed by the following equations. Solvethem by means of Laplace transforms.

du1

dt= −β1(u1 − u2),

du2

dt= −β2(u2 − u1),

u1(0) = 1, u2(0) = 0.

5. Solve the following nonhomogeneous problem with transforms.

∂2u

∂x2= ∂u

∂t− 1, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = 0, 0 < x < 1.

6. Find the transform of the solution of the problem

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 1, 0 < t,

u(x,0) = 0, 0 < x < 1.

7. Find the solution of the problem in Exercise 6 by using the extendedHeaviside formula.

8. Solve the heat problem

∂2u

∂x2= ∂u

∂t, 0 < x, 0 < t,

u(0, t) = 0, 0 < t,

u(x,0) = sin(x), 0 < x.


9. Find the transform of the solution of

∂2u

∂x2= ∂u

∂t, 0 < x, 0 < t,

u(0, t) = 0, 0 < t,

u(x,0) = 1, 0 < x,

u(x, t) bounded as x → ∞.

10. At the end of Section 2.12, the problem in Exercise 9 was solved by othermeans. Use this fact to identify

1

s

(1 − e−√

sx) =L

[erf

(x√4t

)]

and

1

se−√

sx =L[

1 − erf

(x√4t

)].

(The latter function is called the complementary error function, definedby erfc(q) ≡ 1 − erf(q).)

11. Find the function of t whose Laplace transform is

F(s) = e−x√

s.

12. Using the definition of sinh in terms of exponentials and a geometricseries, show that

sinh(√

sx)

sinh(√

s)=

∞∑n=0

(e−√

s(2n+1−x) − e−√s(2n+1+x)

).

13. Use the series in Exercise 12 to find a solution of the problem in Exer-cise 6 in terms of complementary error functions.

14. Show the following relation by using Exercise 11 and differentiating withrespect to s.

L[

1√π t

exp

(−k2

4t

)]= 1√

se−k

√s.

15. Find the Laplace transform of the odd periodic extension of the function

f (t) = π − t, 0 < t < π,

by transforming its Fourier series term by term.


16. Suppose that the function f (t) is periodic with period 2a. Show that theLaplace transform of f is given by the formula

F(s) = G(s)

1 − e−2as,

where

G(s) =∫ 2a

0f (t)e−st dt.

(Hint: See Section 6.1, Exercise 6.)

17. Apply the extended Heaviside method to the inversion of a transformwith the form

F(s) = G(s)

1 − e−2as,

where G(s) does not become infinite for any value of s.

18. Show that for a periodic function f (t) the quantities

cn = 1

2aG

(inπ

a

)

(G(s) is defined in Exercise 16) are the complex Fourier coefficients.

19. How is it possible to determine that a Laplace transform F(s) corre-sponds to a periodic f (t)?

20. Is this function the transform of a periodic function?

F(s) = 1

s2 + a2.

21. Use the method of Exercise 16 to find the transform of the periodic ex-tension of

f (t) ={1, 0 < t < π ,

−1, π < t < 2π .

22. Same as Exercise 21, but use the function of Exercise 15.

23. Use the method of Exercise 16 to find the transform of

f (t) = ∣∣sin(t)∣∣.

24. Find the transform of the solution of the problem


∂2u

∂x2= ∂2u

∂t2, 0 < x, 0 < t,

u(0, t) = h(t), 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 0, 0 < x,

u(x, t) bounded as x → ∞.

Use the solution of the same problem as found in Section 3.6, to verifythe rule

L−1(e−sxH(s)

) ={

h(t − x), t > x,0, t < x.

25. Solve this wave problem with time-varying boundary condition, assum-ing ω �= nπ , n = 1,2, . . . .

∂2u

∂x2= ∂2u

∂t2, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = sin(ωt), 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 0, 0 < x < 1.

26. Solve the problem in Exercise 25 in the special case ω = π .

27. Certain techniques for growing a crystal from a solution or a melt maycause striations — variations in the concentration of impurities. AuthorsR.T. Gray, M.F. Larrousse, and W.R. Wilcox [Diffusional decay of stria-tions, Journal of Crystal Growth, 92 (1988): 530–542] use a material bal-ance on a slice of a cylindrical ingot to derive this boundary value prob-lem for the impurity concentration, C:

∂

∂x

(D(x)

∂C

∂x

)− V

∂C

∂x= ∂C

∂t, 0 < x, 0 < t,

C(0, t) = Ca + A sin

(2π t

tC

), 0 < t,

C(x,0) = Ca, 0 < x.

Here, V is the crystal growth rate, tC is the striation period, Ca is theaverage concentration in the solid, and D(x) is the diffusivity of the im-purity at distance x from the growth face (which is located at x = 0). Ofcourse, C(x, t) is bounded as x → ∞.


Next, the equations are made dimensionless by introducing new vari-ables:

C = C − Ca

A, x = Vx

D(0), t = V2t

D(0).

The new problem is

∂

∂ x

(D(x)

D(0)

∂C

∂ x

)− ∂C

∂ x= ∂C

∂ t, 0 < x, 0 < t,

C(0, t

) = sin(ωt

), 0 < t,

C(x,0) = 0, 0 < x,

where ω = 2πD(0)/V2tC .

Because D(x) depends in a complicated way on x, a numerical solutionwas used. To check the numerical solution, the authors wished to find ananalytical solution of the problem corresponding to constant diffusivity,D(x) = D(0). Let u be the solution of

∂2u

∂ x2= ∂u

∂ x+ ∂u

∂ t, 0 < x, 0 < t,

u(0, t

) = sin(ωt

), 0 < t,

u(x,0) = 0, 0 < x,

u bounded as x → ∞.

Find the Laplace transform of the solution of this problem.

28. The authors of the paper mentioned in Exercise 27 were particularly in-terested in the persistent part of the solution. Use the methods of Sec-tion 6.4 to show that the persistent part of the solution is

u1 = 1

2i

(f (iω) − f (−iω)

),

where

f (iω) = exp

((1

2−

√1

4+ iω

)x + iωt

).

29. Find the square root required in the foregoing expression by setting

√1

4+ iω = α + iβ


so that

α2 − β2 + 2iαβ = 1

4+ iω,

or {α2 − β2 = 1

4,

2αβ = ω.

(To solve these equations: (i) solve the second for β ; (ii) substitute theexpression found into the first; (iii) solve the resulting biquadratic for α.)

30. Noting that the persistent part is u1 = Im(f (iω)) (see Exercise 28), de-termine

u1

(x, t

) = e( 12 −α)x sin

(ωt − β x

),

where α and β are as in Exercise 29.

Numerical MethodsC H A P T E R

7

7.1 Boundary Value Problems

More often than not, significant practical problems in partial — and even or-dinary — differential equations cannot be solved by analytical methods. Dif-ficulties may arise from variable coefficients, irregular regions, unsuitableboundary conditions, interfaces, or just overwhelming detail. Now that ma-chine computation is cheap and easily accessible, numerical methods providereliable answers to formerly difficult problems. In this chapter we examinea few methods that are simple and equally adaptable to machine or manualcomputation. Implementation of some of these methods with a spreadsheetprogram is explained and carried out on the CD.

If we cannot find a simple analytic formula for the solution of a boundaryvalue problem, we may be satisfied with a table of (approximate) values of thesolution. For instance, the solution of the problem

d2u

dx2− 12xu = −1, 0 < x < 1, (1)

u(0) = 1, u(1) = −1, (2)

may be written out in terms of Airy functions, but the values of u shown inTable 1 are more informative for most of us. One way to obtain such a tableis to replace the original analytical problem by an arithmetical problem, asdescribed in what follows.

397

398 Chapter 7 Numerical Methods

x: 0.0 0.2 0.4 0.6 0.8 1.0u(x): 1.0 0.643 0.302 −0.026 −0.406 −1.0

Table 1 Approximate solution of Eqs. (1) and (2)

Differential equation Boundary conditionu(x) → ui u(0) → u0

d2u

dx2(x) → ui+1 − 2ui + ui−1

(x)2

du

dx(0) → u1 − u−1

2xdu

dx(x) → ui+1 − ui−1

2xu(1) → un

f (x) → f (xi)du

dx(1) → un+1 − un−1

2x

Table 2 Constructing replacement equations

First, the values of x for the table will be uniformly spaced across the interval0 ≤ x ≤ 1, which we assume to be the interval of the boundary value problem

xi = ix, x = 1

n.

These are called meshpoints. The numbers approximating the values of u are

ui∼= u(xi), i = 0,1, . . . ,n.

These numbers are required to satisfy a set of equations obtained from theboundary value problem by making the replacements shown in Table 2. Theentry f (x) refers to any coefficient or inhomogeneity in the differential equa-tion.

Example.The boundary value problem in Eqs. (1) and (2) would be replaced by thealgebraic equations

ui+1 − 2ui + ui−1

(x)2− 12xiui = −1, i = 1,2, . . . ,n − 1, (3)

u0 = 1, un = −1. (4)

Equation (3) holds for i = 1, . . . ,n − 1, so the unknowns u1, . . . ,un−1 wouldbe determined by this set of equations. The equations become specific when wechoose n. Let us take n = 5, so x = 1/5, and the four (i = 1,2,3,4) versions

Chapter 7 Numerical Methods 399

of Eq. (3) are

25(u2 − 2u1 + u0) − 12

5u1 = −1,

25(u3 − 2u2 + u1) − 24

5u2 = −1,

25(u4 − 2u3 + u2) − 36

5u3 = −1,

25(u5 − 2u4 + u3) − 48

5u4 = −1.

(5)

When we use the boundary conditions

u0 = 1, u5 = −1 (6)

and collect coefficients, the foregoing equations become

−52.4u1 + 25u2 = −2625u1 − 54.8u2 + 25u3 = −1

25u2 − 57.2u3 + 25u4 = −125u3 − 59.6u4 = 24.

(7)

This system of four simultaneous equations can be solved manually by elimi-nation or by software. The result will be a set of numbers giving the approxi-mate values of u at the points x1 = 0.2, . . . , x4 = 0.8. The numbers in Table 1were obtained by a similar process, but using n = 100 instead of n = 5. �

Example.To see how to handle derivative boundary conditions, we solve the problem

d2u

dx2− 10u = f (x), 0 < x < 1, (8)

u(0) = 1,du

dx(1) = −1, (9)

f (x) =

0, 0 < x < 12 ,

−50, x = 12 ,

−100, 12 < x < 1.

The replacement equations for this problem are easily obtained by using Ta-ble 2. They are

ui+1 − 2ui + ui−1

(x)2− 10ui = f (xi), (10)

u0 = 1,un+1 − un−1

2x= −1. (11)


We need to know u0,u1, . . . ,un. The derivative boundary condition at x = 1forces us to include un+1 among the unknowns, so we will need to use Eq. (10)for i = 1,2, . . . ,n in order to have enough equations to find all the unknowns.Since we have no use for un+1, the usual practice is to solve the boundary-condition replacement for un+1,

un+1 = un−1 − 2x, (12)

and then to use this expression in the version of Eq. (10) that corresponds toi = n. Thus, the equation

un+1 − 2un + un−1

(x)2− 10un = f (xn)

is combined with Eq. (12) to get

2un−1 − 2x − 2un

(x)2− 10un = f (xn). (13)

Then Eq. (10) for i = 1, . . . ,n−1 and Eq. (13) give n equations that determineunknowns u1,u2, . . . ,un.

To be specific, let us take n = 4, so x = 1/4. The three (i = 1,2,3) versionsof Eq. (10) are

16(u2 − 2u1 + u0) − 10u1 = 0 (i = 1),

16(u3 − 2u2 + u1) − 10u2 = −50 (i = 2),

16(u4 − 2u3 + u2) − 10u3 = −100 (i = 3)

and Eq. (13) adapted to n = 4 is

16

(2u3 − 1

2− 2u4

)− 10u4 = −100.

When these equations are cleaned up and the boundary condition u0 = 1 isapplied, the result is the following system of four equations:

−42u1 + 16u2 = −1616u1 − 42u2 + 16u3 = −50

16u2 − 42u3 + 16u4 = −10032u3 − 42u4 = −92.

(14)

In Table 3 are shown the values of ui obtained by solving Eq. (14) and alsomore exact values found by using n = 100. �

Elimination is not the only way to get the solution of a system like Eqs. (7)or (14). An alternative is an iterative method, which generates a sequence ofapproximate solutions. For one such method, we solve algebraically the ith

Chapter 7 Numerical Methods 401

x: 0 0.25 0.5 0.75 1u (n = 4): 1 2.174 4.707 7.057 7.567u (n = 100): 1 2.155 4.729 7.125 7.629

Table 3 Approximate solution of Eqs. (8) and (9)

equation for the ith unknown. In the resulting set of equations there are “cir-cular references”: The equation for u2 refers to u1 and u3, while the equationsfor these refer to u2, etc. We may start with some guessed values for the u’s,feed them through the equations to get improved values for the u’s, and repeatthe process until the values settle down. This method requires a lot of arith-metic but no strategy, while elimination is just the reverse. It may also workwith nonlinear equations, where elimination cannot.

So far, we have given no justification for the procedure of constructing re-placement equations. The explanation is not difficult; it depends on the factthat certain difference quotients approximate derivatives. If u(x) is a functionwith several derivatives, then

u(xi+1) − u(xi−1)

2x= u′(xi) + (x)2

6u(3)(xi), (15)

u(xi+1) − 2u(xi) + u(xi−1)

(x)2= u′′(xi) + (x)2

12u(4)( ¯xi), (16)

where xi and ¯xi are points near xi.Now suppose that u(x) is the solution of the boundary value problem

d2u

dx2+ k(x)

du

dx+ p(x)u(x) = f (x), 0 < x < 1, (17)

αu(0) − α′u′(0) = a, βu(1) + β ′u′(1) = b. (18)

If u(x) has enough derivatives, then at any point xi = ix it satisfies the dif-ferential equation (17) and thus also satisfies the equation

u(xi+1) − 2u(xi) + u(xi−1)

(x)2+ k(xi)

u(xi+1) − u(xi−1)

2x+ p(xi)u(xi) = f (xi)+ δi,

(19)where

δi = (x)2

12u(4)( ¯xi) + k(xi)

(x)2

6u(3)(xi).

Because δi is proportional to (x)2, it is very small when x is small.The replacement equation for Eq. (17) is, according to Table 2,

ui+1 − 2ui + ui−1

(x)2+ k(xi)

ui+1 − ui−1

2x+ p(xi)ui = f (xi). (20)


Thus, the values of u at x0, x1, . . . , xn, which satisfy Eq. (19) exactly, will nearlysatisfy Eq. (20); vice versa, the numbers u0,u1, . . . ,un, which satisfy the re-placement equations (20), nearly satisfy Eq. (19). It can be proved that thecalculated numbers u0,u1, . . . ,un do indeed approach the appropriate valuesof u(xi) as x approaches 0 (under continuity and other conditions on k(x),p(x), f (x)).

E X E R C I S E S

1. Set up and solve replacement equations with n = 4 for the problem

d2u

dx2= −1, 0 < x < 1,

u(0) = 0, u(1) = 1.

2. Solve the problem of Exercise 1 analytically. On the basis of Eqs. (15)and (16), explain why the numerical solution agrees exactly with the ana-lytical solution.


d2u

dx2− u = −2x, 0 < x < 1,

u(0) = 0, u(1) = 1.

4. Solve the problem in Exercise 3 analytically, and compare the numericalresults with the true solution.


d2u

dx2= x, 0 < x < 1,

u(0) − du

dx(0) = 1, u(1) = 0.

6. Solve the problem in Exercise 5 analytically, and compare the numericalresults with the true solution.

7. Set up and solve replacement equations for the problem

d2u

dx2+ 10u = 0, 0 < x < 1,

u(0) = 0, u(1) = −1.

7.2 Heat Problems 403

Use n = 3 and n = 4. Sketch the results and explain why they vary somuch.

In Exercises 8–11, set up and solve replacement equations for the problemstated and the given value of n. If a computer is available, also solve for n twiceas large, and compare results.

8.d2u

dx2− 32xu = 0, 0 < x < 1,

u(0) = 0, u(1) = 1 (n = 4).

9.d2u

dx2− 25u = −25, 0 < x < 1,

u(0) = 2, u(1) + u′(1) = 1 (n = 5).

10.d2u

dx2+ 1

1 + x

du

dx= −1, 0 < x < 1,

u(0) = 0, u(1) = 0 (n = 3).

11.d2u

dx2+ du

dx− u = −x,

du

dx(0) = 0, u(1) = 1 (n = 3).

12. Use the Taylor series expansion

u(x + h) = u(x) + hu′(x) + h2

2u′′(x) + h3

6u(3)(x) + h4

24u(4)(x) + · · ·

with x = xi and h = ±x (xi + x = xi+1, xi − x = xi−1) to obtain rep-resentations similar to Eqs. (15) and (16).

7.2 Heat ProblemsIn heat problems, we have two independent variables x and t, assumed to bein the range 0 < x < 1, 0 < t. A table for a function u(x, t) should give valuesat equally spaced points and times,

xi = ix, tm = mt,

for i = 0,1, . . . ,n and m = 0,1, . . . . Here, x = 1/n, as before. We will use asubscript to denote position and a number in parentheses to denote the timelevel for the approximation to the solution of a problem. That is,

ui(m) ∼= u(xi, tm).


The spatial derivatives in a heat problem will be replaced by difference quo-tients, as before:

∂2u

∂x2(xi, tm) → ui+1(m) − 2ui(m) + ui−1(m)

(x)2, (1)

∂u

∂x(xi, tm) → ui+1(m) − ui−1(m)

2x. (2)

For the time derivative, there are several possible replacements. We limit our-selves to the forward difference

∂u

∂t(xi, tm) → ui(m + 1) − ui(m)

t, (3)

which will yield explicit formulas for computing.Now, to solve numerically the simple heat problem

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t, (4)

u(0, t) = 0, u(1, t) = 0, 0 < t, (5)

u(x,0) = f (x), 0 < x < 1, (6)

we set up replacement equations according to Eqs. (1)–(3). Those equationsare

ui−1(m) − 2ui(m) + ui+1(m)

(x)2= ui(m + 1) − ui(m)

t, (7)

supposed valid for i = 1,2, . . . ,n − 1 and m = 0,1,2, . . . .The point of using a forward difference for the time derivative is that these

equations may be solved for ui(m + 1):

ui(m + 1) = rui−1(m) + (1 − 2r)ui(m) + rui+1(m), (8)

where r = t/(x)2. Thus each ui(m + 1) is calculated from u’s at the preced-ing time level. Because the initial condition gives each ui(0), the values of theu’s at time level 1 can be calculated by setting m = 0 in Eq. (8):

ui(1) = rui−1(0) + (1 − 2r)ui(0) + rui+1(0).

Then the values of the u’s at time level 2 can be found from these, and so oninto the future. Of course, r has to be given a numerical value first, by choosingx and t.

It is convenient to display the numerical values of ui(m) in a table, makingcolumns correspond to different meshpoints x0, x1, . . . , xn and making rowscorrespond to the different time levels t0, t1, . . . . See Table 4.


i

m 0 1 2 3 40 0 0.25 0.5 0.75 11 0 0.25 0.5 0.75 02 0 0.25 0.5 0.25 03 0 0.25 0.25 0.25 04 0 0.125 0.25 0.125 05 0 0.125 0.125 0.125 0

Table 4 Numerical solution of Eqs. (4)–(6)

Example.Solve Eqs. (4)–(6) with x = 1/4 and r = 1/2, making t = 1/32. The equa-tions giving the u’s at time level m + 1 are

u1(m + 1) = 1

2

(u0(m) + u2(m)

),

u2(m + 1) = 1

2

(u1(m) + u3(m)

),

u3(m + 1) = 1

2

(u2(m) + u4(m)

).

(9)

Recall that the boundary conditions of this problem specify u0(m) = 0 andu4(m) = 0 for m = 1,2,3, . . . . Thus we fill in the columns of the table thatcorrespond to points x0 and x4 with 0’s (shown in italics in Table 4). Alsothe initial condition specifies ui(0) = f (xi), so the top row of the table can befilled. In this example we take f (x) = x, and the corresponding values appearin italics in the top row of Table 4.

The initial condition, u(x,0) = x, 0 < x < 1, suggests that u(1,0) shouldbe 1, while the boundary condition suggests that it should be 0. In fact, nei-ther condition specifies u(1,0), nor is there a hard and fast rule telling what todo in case of conflict. Fortunately, it does not matter much, either. (See Exer-cise 1.) �

StabilityThe choice we made of r = 1/2 in the Example seems natural, perhaps, becauseit simplifies the computation. It might also seem desirable to take a larger valueof r (signifying a larger time step) to get into the future more rapidly. Forexample, with r = 1 (t = 1/16) the replacement equations take the form

ui(m + 1) = ui−1(m) − ui(m) + ui+1(m).

In Table 5 are values of ui(m) computed from this formula. No one can believethat these wildly fluctuating values approximate the solution to the heat prob-


i

m 0 1 2 3 40 0 0.25 0.50 0.75 11 0 0.25 0.50 0.75 02 0 0.25 0.50 −0.25 03 0 0.25 −0.50 0.75 04 0 −0.75 1.50 −1.25 05 0 2.25 −3.50 2.75 0

Table 5 Unstable solution

lem in any sense. Indeed, they suffer from numerical instability due to using atime step too long relative to the mesh size. The analysis of instability requiresfamiliarity with matrix theory, but there are some simple rules of thumb thatguarantee stability.

First, write out the equations for each ui(m + 1):

ui(m + 1) = aiui−1(m) + biui(m) + ciui+1(m).

The coefficients must satisfy two conditions

1. No coefficient may be negative.

2. The sum of the coefficients is not greater than 1.

In the example, the replacement equations were

u1(m + 1) = ru0(m) + (1 − 2r)u1(m) + ru2(m),

u2(m + 1) = ru1(m) + (1 − 2r)u2(m) + ru3(m),

u3(m + 1) = ru2(m) + (1 − 2r)u3(m) + ru4(m).

The second requirement is satisfied automatically, because r+(1−2r)+ r = 1.But the first condition is satisfied only for r ≤ 1/2. Thus the first choice ofr = 1/2 corresponded to the longest stable time step.

Example.Different problems give different maximum values for r. For the heat conduc-tion problem

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t, (10)

u(0, t) = 1,∂u

∂x(1, t) + γ u(1, t) = 0, 0 < t, (11)

u(x,0) = 0, 0 < x < 1 (12)


the replacement equations are found to be (for n = 4)

u1(m + 1) = ru0(m) + (1 − 2r)u1(m) + ru2(m),

u2(m + 1) = ru1(m) + (1 − 2r)u2(m) + ru3(m),

u3(m + 1) = ru2(m) + (1 − 2r)u3(m) + ru4(m),

u4(m + 1) = 2ru3(m) + (1 − 2r − 1

2 rγ)u4(m).

(13)

(Remember that u(1, t), corresponding to u4, is an unknown. The boundarycondition has been incorporated into the equation for u4(m + 1).) Again, thesecond stability requirement is satisfied automatically; but the first rule re-quires that

1 − 2r − 1

2rγ ≥ 0 or r ≤ 1

2 + 12γ

. (14)

�

E X E R C I S E S

1. Solve Eqs. (4)–(6) numerically with f (x) = x, as in the text (x = 1/4,r = 1/2), but take u4(0) = 0. Compare your results with Table 4.

2. Solve Eqs. (4)–(6) numerically with f (x) = x, x = 1/4, u4(0) = 1, as inthe text, but use r = 1/4. Compare your results with Table 4. Be sure tocompare results at corresponding times.

3. For the problem in Eqs. (10)–(12), find the longest stable time step whenγ = 1, and compute the numerical solution with the corresponding valueof r.

4. Solve the problem in Eqs. (10)–(12) with x = 1/4, r = 1/2 and γ = 0, form up to 5.

For each problem in the following exercises, set up the replacement equationsfor n = 4, compute the longest stable time step, and calculate the numericalsolution for a few values of m.

5.∂2u

∂x2= ∂u

∂t, u(0, t) = u(1, t) = t, u(x,0) = 0.

6.∂2u

∂x2− u = ∂u

∂t, u(0, t) = u(1, t) = 1, u(x,0) = 0.

7.∂2u

∂x2= ∂u

∂t− 1, u(0, t) = u(1, t) = 0, u(x,0) = 0.


8.∂2u

∂x2= ∂u

∂t, u(0, t) = 0,

∂u

∂x(1, t) + u(1, t) = 1, u(x,0) = 0.

9.∂2u

∂x2= ∂u

∂t,

∂u

∂x(0, t) = 0, u(1, t) = 1, u(x,0) = x.

7.3 Wave EquationThe simple vibrating string problem we studied in Chapter 3,

∂2u

∂x2= ∂2u

∂t2, 0 < x < 1, 0 < t, (1)

u(0, t) = 0, u(1, t) = 0, 0 < t, (2)

u(x,0) = f (x),∂u

∂t(x,0) = g(x), 0 < x < 1, (3)

rarely needs treatment by numerical methods, because the d’Alembert solu-tion provides a simple and direct means of calculating the solution u(x, t) forarbitrary x and t. However, if the partial differential equation contains u or aninhomogeneity or if the boundary conditions are more complex, a series solu-tion or a solution of the d’Alembert type may not be practical. In many suchcases, simple numerical techniques are quite rewarding.

In order to convert the wave equation (1) into a suitable difference equa-tion, we first designate points xi = ix(x = 1/n) and times tm = mt forwhich the approximation to u will be found: u(xi, tm) ∼= ui(m). Then the par-tial derivatives with respect to both x and t are replaced by central differences:

∂2u

∂x2→ ui+1(m) − 2ui(m) + ui−1(m)

(x)2,

∂2u

∂t2→ ui(m + 1) − 2ui(m) + ui(m − 1)

(t)2.

The wave equation (1) becomes this partial difference equation

ui+1(m) − 2ui(m) + ui−1(m)

(x)2= ui(m + 1) − 2ui(m) + ui(m − 1)

(t)2

or, with ρ = t/x,

ui(m + 1) − 2ui(m) + ui(m − 1) = ρ2(ui+1(m) − 2ui(m) + ui−1(m)

).

The replacement equations may be solved for the unknowns ui(m + 1),yielding the equation

ui(m + 1) = ρ2ui−1(m) + 2(1 − ρ2)ui(m) + ρ2ui+1(m) − ui(m − 1), (4)

7.3 Wave Equation 409

valid for i = 1,2, . . . ,n − 1. Naturally, the boundary conditions, Eq. (2), carryover as u0(m) = 0, un(m) = 0. It is obvious that Eq. (4) requires us to know theapproximate solution at time levels m and m−1 in order to find it at time levelm + 1. In other words, to get ui(1) we need ui−1(0), ui(0), ui+1(0) — which areavailable from the initial condition — and also ui(−1)! Of course, we have notyet applied the second initial condition,

∂u

∂t(x,0) = g(x), 0 < x < 1.

If we replace the time derivative by a central difference approximation, thisequation translates into

ui(1) − ui(−1)

2t= g(xi) (5)

for i = 1,2, . . . ,n − 1. Equation (5), together with a slightly modified versionof Eq. (4) (with m = 0 and ui(0) = f (xi)), yields the system

ui(1) + ui(−1) = ρ2f (xi−1) + 2(1 − ρ2

)f (xi) + ρ2f (xi+1),

ui(1) − ui(−1) = 2t g(xi),(6)

which we can easily solve for the u’s at the first time level:

ui(1) = 1

2ρ2f (xi−1) + (

1 − ρ2)f (xi) + 1

2ρ2f (xi+1) + t g(xi). (7)

Thus, in order to solve the problem in Eqs. (1)–(3) numerically, we use theinitial condition, ui(0) = f (xi), to fill the first line of our table, use the startingequation (7) to fill the next line, and continue with the running equation (4) tofill subsequent lines.

Example.Let us now attempt to solve a simple problem. Suppose that g(x) ≡ 0 for 0 <

x < 1 and that f (x) is given by

f (x) ={

2x, 0 < x < 12 ,

2(1 − x), 12 < x < 1.

(8)

Also, we shall choose n = 4 and ρ = 1 for convenience. (That is, t = x =1/4.) Our rule for calculation, Eq. (4), is then

ui(m + 1) = ui−1(m) + ui+1(m) − ui(m − 1). (9)

In Table 6 are the calculated values of ui(m). Entries in italics are given data.It is easy to check that this numerical solution is identical with the d’Alembertsolution of this particular problem. (See Exercise 6.) However, if the initial


i

m 0 1 2 3 40 0 0.5 1 0.5 01 0 0.5 0.5 0.5 02 0 0 0 0 03 0 −0.5 −0.5 −0.5 04 0 −0.5 −1 −0.5 05 0 −0.5 −0.5 0.5 06 0 0 0 0 0


velocity were not identically zero, the numerical solution would in general beonly an approximation to the true solution. �

StabilityIn our study of the heat equation, Section 7.2, we saw that the choice of x andt was not free. The same is true for the wave equation. Suppose we attemptto solve the same problem as earlier, but with ρ2 = (t/x)2 chosen to be 2.Then Eq. (4) becomes

ui(m + 1) = 2(ui−1(m) − ui(m) + ui+1(m)

) − ui(m − 1),

and the “solution” corresponding to this rule of calculation is shown in Table 7(again, entries in italics are given data). Of course, the results bear no resem-blance to the solution of the wave equation. They suffer from the same sort ofinstability as that observed in Section 7.2. There is a rule of thumb, similar tothe one to be found there, applicable to the wave equation.

First, write out the equations for each ui(m + 1) in terms of the u’s at timelevels m and m − 1:

ui(m + 1) = aiui−1(m) + biui(m) + ciui+1(m) − ui(m − 1).

The coefficients must satisfy two conditions:

1. None of the coefficients ai, bi, ci may be negative.

2. The sum of the coefficients is not greater than 2:

ai + bi + ci ≤ 2.

Of course, ui(m − 1) appears with a coefficient of −1; nothing can be doneabout that, nor does it enter into the aforementioned rules.


i

m 0 1 2 3 40 0 0.5 1 0.5 01 0 0.5 0 0.5 02 0 −1.5 1 −1.5 03 0 4.5 −8 4.5 04 0 −23.5 33 −23.5 0

Table 7 Unstable numerical solution

In Eq. (4) we see that both conditions are met when ρ = t/x is less thanor equal to 1; in other words, the time step must not exceed the space step.However, using ρ2 = 1 when acceptable often provides the best accuracy.

We conclude with one more example, illustrating how numerical results canbe obtained easily in some cases that might be puzzling analytically.

Example.Suppose we are to solve the problem

∂2u

∂x2= ∂2u

∂t2− 16 cos(π t), 0 < x < 1, 0 < t, (10)

u(0, t) = 0, u(1, t) = 0, 0 < t, (11)

u(x,0) = 0,∂u

∂t(x,0) = 0, 0 < x < 1. (12)

We replace the partial derivatives as before, obtaining

ui+1(m) − 2ui(m) + ui−1(m)

(x)2

= ui(m + 1) − 2ui(m) + ui(m − 1)

(t)2− 16 cos(π tm).

When this is solved for ui(m + 1), we find

ui(m + 1) = (2 − 2ρ2)ui(m) + ρ2ui+1(m) + ρ2ui−1(m)

− ui(m − 1) + 16(t)2 cos(πmt). (13)

Let us take x = t = 1/4 again, so ρ = 1 and Eq. (13) simplifies to

ui(m + 1) = ui+1(m) + ui−1(m) − ui(m − 1) + cos

(mπ

4

). (14)

This is our running equation. The starting equation comes from combiningEqs. (14) for m = 0,

ui(1) = −ui(−1) + 1


i

m 0 1 2 3 40 0 0 0 0 01 0 0.5 0.5 0.5 02 0 1.21 1.71 1.21 03 0 1.21 1.91 1.21 04 0 0.00 0.00 0.00 05 0 −2.21 −2.91 −2.21 06 0 −3.62 −5.12 −3.62 07 0 −2.91 −4.33 −2.91 0


(note ui(0) = 0), with the replacement initial condition

ui(1) − ui(−1)

2t= 0,

or

ui(1) = ui(−1) = 1

2for i = 1,2,3. Now we have the top two lines of Table 8, and the rest are filledusing Eqs. (14) (with cos(π/4) � 0.71, and so forth). Entries in italics are givendata.

The complete analytical solution of this problem is

u(x, t) = 32

π2t sin(π t) sin(πx)

+ 32

π3

∞∑n=3

1 − cos(nπ)

n(n2 − 1)

(cos(π t) − cos(nπ t)

)sin(nπx).

At x = 1/2, the sum of the infinite series is 0, so

u

(1

2, t

)= 32

π2t sin(π t).

Comparison of the values of this function at times tm with the middle columnof Table 8 shows the numerical solution off by a few percent. Note that thegrowth in u(x, t) is due to resonance in the physical system, not to numericalinstability. �

E X E R C I S E S

1. Obtain an approximate solution of Eqs. (1), (2), and (3) with f (x) ≡ 0 andg(x) ≡ 1. Take x = 1/4, ρ = 1.


2. Compare the results of Exercise 1 with the d’Alembert solution.

3. Obtain an approximate solution of Eqs. (1), (2), and (3) with f (x) ≡ 0 andg(x) = sin(πx). Take x = 1/4, ρ = 1.

4. Compare the results of Exercise 3 with the exact solution u(x, t) =(1/π) sin(πx) sin(π t).

5. Obtain an approximate solution of Eqs. (1), (2), and (3) with g(x) ≡ 0 andf (x) as in Eq. (8). Use x = 1/4 and ρ2 = 1/2.

6. Compare the entries of Table 6 with the d’Alembert solution.

7. Obtain an approximate solution of this problem with a time-varyingboundary condition, using x = t = 1/4.

∂2u

∂x2= ∂2u

∂t2, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = h(t), 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 0, 0 < x < 1,

h(t) ={1, 0 < t < 1,

−1, 1 < t < 2

and h(t + 2) = h(t), h(0) = h(1) = 0.

8. Same task as Exercise 7 but h(t) = sin(π t). Use sin(π/4) ∼= 0.7 instead of√2/2.

9. Find starting and running equations for the following problem. Usingx = 1/4, find the longest stable time step and compute values of theapproximate solution for m up to 8.

∂2u

∂x2= ∂2u

∂t2+ 16u, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = f (x),∂u

∂t(x,0) = 0, 0 < x < 1,

where f (x) is given in Eq. (8).

10. Using x = 1/4 and ρ2 = 1/2, compare the numerical solution of theproblem in Exercise 9 with and without the 16u term in the partial differ-ential equation.


7.4 Potential EquationIn this section, we will be concerned with approximate solutions of the po-tential equation and related equations in a region R of the xy-plane. For thesake of simplicity, we will limit ourselves to regions whose boundaries can bemade to coincide with the lines on a sheet of graph paper with square divi-sions. Thus, we admit such shapes as rectangles, L’s and T’s, but not circles ortriangles. The graph paper provides us with a ready-made mesh of points inthe region R and on its boundary, at which we wish to know the solution ofour problem. These points are to be numbered in some fashion — usually leftto right and bottom to top.

On such a mesh, the replacement for the Laplacian operator is the following:

∂2u

∂x2+ ∂2u

∂y2→ uW − 2ui + uE

(x)2+ uN − 2ui + uS

(y)2, (1)

where the subscripts E, W stand for the indices of the mesh points to the leftand right of point i and the subscripts N , S stand for those above and below(see Fig. 1). The result is sometimes called the five-point approximation to theLaplacian. Because we are assuming that x = y, we obtain a further sim-plification in the replacement:

∂2u

∂x2+ ∂2u

∂y2→ uN + uS + uE + uW − 4ui

(x)2. (2)

Example.Solve this problem numerically (see Chapter 4 for the analytical solution):

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < 1, 0 < y < 1, (3)

u(0, y) = 0, u(1, y) = 0, 0 < y < 1, (4)

u(x,0) = f (x), u(x,1) = f (x), 0 < x < 1, (5)

f (x) ={

2x, 0 < x < 12 ,

2(1 − x), 12 ≤ x < 1.

(6)

Let us take x = y = 1/4 and number the mesh points inside the 1 × 1square as shown in Fig. 2.

At each of the nine mesh points, we will have the replacement equation

uN + uS + uE + uW − 4ui = 0. (7)

Together, these make up a system of nine equations in the nine unknownsu1,u2, . . . ,u9. Referring to Fig. 2, where the values of u at boundary points are

7.4 Potential Equation 415

Figure 1 Point i on a square mesh and its four neighbors.

Figure 2 Numbering for mesh points, and values on boundary.

shown, we can write down the equations to be solved:

u2 + u4 + 12 − 4u1 = 0

u1 + u3 + u5 + 1 − 4u2 = 0

u2 + u6 + 12 − 4u3 = 0

u1 + u5 + u7 − 4u4 = 0

u2 + u4 + u6 + u8 − 4u5 = 0

u3 + u5 + u9 − 4u6 = 0

u4 + u8 + 12 − 4u7 = 0

u5 + u7 + u9 + 1 − 4u8 = 0

u6 + u8 + 12 − 4u9 = 0.

(8)

This is simply a system of simultaneous equations. It can be solved by elim-ination to obtain the results shown in Fig. 3. In this particular case, there arenumerous symmetries in the problem, so u1 = u3 = u7 = u9, u2 = u8, andu4 = u6. Thus, only u1,u2,u4, and u5 need to be found. The system can bereduced to four equations in these four unknowns, which can even be solvedmanually. �


Figure 3 Numerical solution of Eqs. (3)–(6).

Example.Set up the replacement equations for the problem

∂2u

∂x2+ ∂2u

∂y2= 16(u − 1), 0 < x < 1, 0 < y < 1, (9)

u(x,0) = 0, u(x,1) = 0, 0 < x < 1, (10)

u(0, y) = 0, u(1, y) = 0, 0 < y < 1. (11)

We may use the same numbering as in the first example (Fig. 2). At each meshpoint, the replacement is

uN + uS + uE + uW − 4ui

(x)2= 16(ui − 1). (12)

Because x = 1/4, (1/x)2 = 16, and the typical replacement equation be-comes

uN + uS + uE + uW − 4ui = ui − 1,

or

uN + uS + uE + uW − 5ui = −1. (13)

Finally, we may write out the equations to be solved. The first four of the nineequations, corresponding to Eq. (13) with i = 1,2,3,4, are

u2 + u4 − 5u1 = −1u1 + u3 + u5 − 5u2 = −1

u2 + u6 − 5u3 = −1u1 + u5 + u7 − 5u4 = −1.

(14)

The solution of this problem is left as an exercise. �

On more complicated regions, the replacement for the Laplacian operatorhas exactly the same form, since we still use the “graph-paper mesh.” The sys-


Figure 4 Mesh numbering for L-shaped region.

tem of equations to be solved will be rather less regular than that for a rectan-gle.

Example.Consider the problem

∂2u

∂x2+ ∂2u

∂y2= −16 in R, (15)

u = 0 on the boundary of R, (16)

whereR is an L-shaped region formed from a 1×1 square by removing a 1/4×1/4 square from the upper right corner. The general replacement equation is

uN + uS + uE + uW − 4ui = −1. (17)

With the numbering shown in Fig. 4, the eight equations to be solved are

u2 + u4 − 4u1 = −1u1 + u3 + u5 − 4u2 = −1

u2 + u6 − 4u3 = −1u1 + u5 + u7 − 4u4 = −1

u2 + u4 + u6 + u8 − 4u5 = −1u3 + u5 − 4u6 = −1u4 + u8 − 4u7 = −1u5 + u7 − 4u8 = −1.

(18)

The results, rounded to three digits, are shown in Eq. (19). Note the equalities,which arise from symmetries in the problem:

u1 = 0.656u2 = u4 = 0.813u3 = u7 = 0.616

u5 = 0.981u6 = u8 = 0.649.

(19)

�


Iterative MethodsSystems of up to 10 equations, such as those in the foregoing examples, canreadily be solved by elimination. It is easy to see, however, that we might wellwant a finer mesh to get better accuracy and that a finer mesh will increase thenumber of equations dramatically. For example, if we use x = y = 1/10in a numerical solution of Eqs. (3)–(5), the system to be solved contains 81unknowns (or 25 if we use symmetry). Problems involving many thousands ofunknowns are quite common. These large systems of simultaneous equationsare almost always solved by iterative methods, which generate a sequence ofapproximate solutions.

Consider again the potential problem in Eqs. (3)–(6). Let us take a meshwith x = y = 1/N and number the points of the mesh with a double indexso that

u(xi, yj) ∼= ui,j. (20)

Then the replacement equations for the potential equation are

ui+1,j − 2ui,j + ui−1,j

(x)2+ ui,j+1 − 2ui,j + ui,j−1

(y)2= 0,

or, using x = y and some algebra,

ui,j = 1

4(ui+1,j + ui−1,j + ui,j+1 + ui,j−1), (21)

valid for i and j ranging from 1 to N − 1. (This is the same as Eq. (7).) Theboundary conditions, Eqs. (4) and (5), determine

u0,j = 0, uN,j = 0, j = 0, . . . ,N, (22)

ui,0 = f (xi), ui,N = f (xi), i = 0, . . . ,N. (23)

The simplest iterative method, called the Gauss–Seidel method, works thisway. We sweep through the array of u’s, replacing each ui,j by the combinationof u’s given on the right-hand side of Eq. (21). After several sweeps throughthe array, the numbers no longer change much. When the new and old valuesof ui,j at each point agree closely enough, we stop.

The result is a set of numbers that satisfy Eq. (21) approximately. Since theexact solution of the replacement equations is still just an approximation tothe solution of the original problem in Eqs. (3)–(6), it is not urgent to get thatexact solution of the replacement equations.

An iterative method such as the Gauss–Seidel method is very easy to imple-ment on a spreadsheet without programming. (See the CD.)


Figure 5 Regions and mesh numbering for Exercises 5–9.

E X E R C I S E S

Set up and solve replacement equations for each of the following problems.Use symmetry to reduce the number of unknowns.1. ∇2u = −1, 0 < x < 1, 0 < y < 1, u = 0 on the boundary. x = y = 1/4.

2. Same as Exercise 1 with x = y = 1/8. Compare the solutions.

3. ∇2u = 0, 0 < x < 1, 0 < y < 1, u(0, y) = 0, u(x,0) = 0, u(1, y) = y,u(x,1) = x. x = y = 1/4.

4. Same as Exercise 3 with x = y = 1/8.

5. The region R is a square of side 1 from the center of which a similar squareof side 1/7 has been removed; ∇2u = 0 in R, u = 0 on the outside bound-ary, and u = 1 on the inside boundary; x = y = 1/7. See Fig. 5.

6. Same as Exercise 5, but the partial differential equation is ∇2u = −1, andthe boundary condition is u = 0 on all boundaries. See Fig. 5.

7. The region R has the shape of a T, made by removing strips from the cor-ners of a 1 × 1 square. The partial differential equation is ∇2u = −25 in R,and u = 0 on the boundary. Take x = y = 1/5. See Fig. 5 for numberingof mesh points.

8. The region is a rectangle, 2 units wide and 1 unit high. The potential equa-tion holds in the interior; u = 1 on the upper half of the boundary (the top


and the upper halves of the vertical sides), and u = 0 on the lower half. Takex = y = 1/3. See Fig. 5.

9. The region, as seen in Fig. 5, is shaped like an upside-down U and is formedby removing a small (1 × 2) rectangle from the bottom of a larger (5 × 4)

one. In the interior of the region, ∇2u = 0. The boundary conditions are:u = 1 on the left and right sides and the top of the rectangle; u = 0 on thebottom and on the boundary formed by the removal of the small rectangle.Use x = y = 1.

7.5 Two-Dimensional ProblemsSeparation of variables and other analytical methods produce satisfactory so-lutions to two-dimensional problems in only the nicest cases. However, simplenumerical methods work quite well on two-dimensional problems. In this ele-mentary exposition, we will limit ourselves to the heat and wave equations ontwo-dimensional regions that “fit on graph paper,” as in Section 7.4.

We will compute an approximation to the solution of a problem, denot-ing space position with one or two subscripts and time level with an index inparentheses. Both heat and wave problems will require the replacement of theLaplacian operator. We use the same replacement as in Section 7.4,

∂2u

∂x2+ ∂2u

∂y2→ uE(m) − 2ui(m) + uW(m)

(x)2+ uN(m) − 2ui(m) + uS(m)

(y)2.

Because we are using a square mesh, with x = y, the replacement simplifiesto

∂2u

∂x2+ ∂2u

∂y2→ uN(m) + uS(m) + uE(m) + uW(m) − 4ui(m)

(x)2, (1)

where N , S, E, W stand for the indices of the four grid points adjacent to thepoint with the index i.

Heat ProblemsNow let us consider this heat problem on a rectangle:

∂2u

∂x2+ ∂2u

∂y2= ∂u

∂t, 0 < x < 1.25, 0 < y < 1, 0 < t, (2)

u(0, y, t) = 0, u(1.25, y, t) = 0, 0 < y < 1, 0 < t, (3)

u(x,0, t) = 0, u(x,1, t) = 0, 0 < x < 1.25, 0 < t, (4)

u(x, y,0) = 1, 0 < x < 1.25, 0 < y < 1. (5)

7.5 Two-Dimensional Problems 421

Figure 6 Mesh numbering for numerical solution of Eqs. (2)–(5).

We take x = y = 1/4 and number the interior points of the region asshown in Fig. 6. Then we will be computing the approximations

u1(m) ∼= u

(1

4,

1

4, tm

), u2(m) ∼= u

(1

2,

1

4, tm

), u3(m) ∼= u

(3

4,

1

4, tm

), . . .

(6)and so forth, for m = 1,2, . . . . The replacement equations are obtained usingEq. (1) for the Laplacian and a forward difference to replace the time deriva-tive. The typical equation is

uN(m) + uS(m) + uE(m) + uW(m) − 4ui(m)

(x)2= ui(m + 1) − ui(m)

t. (7)

When we solve this equation for ui(m + 1), we obtain

ui(m + 1) = r[uN(m) + uS(m) + uE(m) + uW(m)] + (1 − 4r)ui(m), (8)

in which

r = t

x2= t

y2= 16t.

The stability considerations of Section 7.2 are still important, and the rules ofthumb are still valid. We must limit r by the requirement that 1 −4r ≥ 0, or, inthis case, t ≤ 1/64. We shall take the longest acceptable time step, t = 1/64,r = 1/4, which makes the equations a little simpler.

At m = 0, all temperatures are given as 1. For m ≥ 1, all the boundary tem-peratures are zero and the ui(m) are all found to equal 1. For m = 2, we calcu-late

u1(2) = 1

4

(u2(1) + u5(1) + 0 + 0

) = 1

2,

u2(2) = 1

4

(u1(1) + u3(1) + u6(1) + 0

) = 3

4,


i

m 1 2 5 60 1 1 1 1

11

2

3

4

3

41

23

8

9

16

1

2

13

16

317

64

7

16

25

64

39

64


...

u5(2) = 1

4

(u1(1) + u6(1) + u9(1) + 0

) = 3

4,

u6(2) = 1

4

(u2(1) + u5(1) + u7(1) + u10(1)

) = 1.

The 0’s in these equations stand for boundary temperatures.An alert calculator will notice that only the unknowns u1,u2,u5,u6 need be

calculated, since, in this example, the others will be given at each time step bysymmetry

u1(m) = u4(m) = u9(m) = u12(m), u5(m) = u8(m),

u6(m) = u7(m), u2(m) = u3(m) = u10(m) = u11(m).

In Table 9 are computed values of the significant u’s at a few times.Now consider this heat problem, which is not solvable by separation of vari-

ables:

∂2u

∂x2+ ∂2u

∂y2= ∂u

∂tin R, (9)

u = f (t) on C, (10)

u = 0 in R at t = 0. (11)

Here, R is an L-shaped region and C is its boundary. The function f we taketo be f (t) = t, but more complicated functions can be used.

To start the numerical solution, we set up a square grid, as shown in Fig. 7.The spacing is x = y = 1/5 and the numbering of the points is shown. Thetypical replacement equation is just as given in Eqs. (7) and (8). We must bearin mind, however, that some points are adjacent to boundary points where thetemperature is given by f (t).


Figure 7 Mesh numbering for numerical solution of Eqs. (9)–(11).

Because x = y = 1/5, the parameter r in Eq. (8) is

r = t

x2= 25t.

Clearly, the longest stable time step is t = 1/100, corresponding to r = 1/4.Using this value of r simplifies the typical replacement equation to

ui(m + 1) = 1

4

(uN(m) + uS(m) + uE(m) + uW(m)

). (12)

Specifically, we have

u1(m + 1) = 1

4

(u2(m) + u5(m) + 2f (tm)

),

u2(m + 1) = 1

4

(u1(m) + u3(m) + u6(m) + f (tm)

),

and so on. The f (tm) terms enter because point 1 is adjacent to two boundarypoints and point 2 to one boundary point. Note that symmetry about the linethrough points 4 and 7 makes it unnecessary to compute u8(m), . . . ,u12(m).Table 10 contains calculated values of u for the first four time levels.

Wave ProblemsIn solving two-dimensional wave problems, we replace the Laplacian, as in theforegoing, and use a central difference for the time derivative, as we did inSection 7.3:

∂2u

∂t2→ ui(m + 1) − 2ui(m) + ui(m − 1)

t2. (13)


i

m 1 2 3 4 5 6 7 f (tm)

0 0 0 0 0 0 0 0 01 0 0 0 0 0 0 0 1.02 0.5 0.25 0.25 0.5 0.5 0.25 0 2.03 1.22 0.75 0.69 0.75 1.22 0.69 0.25 3.04 1.99 1.40 1.19 1.84 1.98 1.30 0.69 4.0

Table 10 Numerical solution of Eqs. (9)–(11). Entries are 100 × ui(m)

As an example, let us consider the vibrations of a square membrane, as de-scribed by the problem

∂2u

∂x2+ ∂2u

∂y2= ∂2u

∂t2, 0 < x < 1, 0 < y < 1, 0 < t, (14)

u(x,0, t) = 0, u(x,1, t) = 0, 0 < x < 1, 0 < t, (15)

u(0, y, t) = 0, u(1, y, t) = 0, 0 < y < 1, 0 < t, (16)

u(x, y,0) = f (x, y), 0 < x < 1, 0 < y < 1, (17)

∂u

∂t(x, y,0) = g(x, y), 0 < x < 1, 0 < y < 1. (18)

A typical replacement for the wave equation (14) is constructed using Eq. (1)for the Laplacian and Eq. (13) for the time derivative:

ui(m + 1) − 2ui(m) + ui(m − 1)

(t)2

= uN(m) + uS(m) + uE(m) + uW(m) − 4ui(m)

(x)2. (19)

As usual we solve for ui(m + 1), using the abbreviation ρ = t/x. The resultis

ui(m + 1) = ρ2[uE(m) + uW(m) + uN(m) + uS(m)

]+ (2 − 4ρ2)ui(m) − ui(m − 1). (20)

The stability rules given earlier still apply. Thus we must choose ρ2 ≤ 1/2 inorder to get a sensible solution.

Let us now be specific. We shall take x = y = 1/4, ρ2 = 1/2 (that is,t = 1/4

√2), and suppose that the initial data from Eqs. (11) and (12) are

f (x, y) ={

1 near x = 14 , y = 1

4 ,0 elsewhere,

g(x, y) ≡ 0.


The running equation is Eq. (20), which, with ρ2 = 1/2, simplifies to

ui(m + 1) = 1

2

[uE(m) + uW(m) + uN(m) + uS(m)

] − ui(m − 1). (21)

To find the starting equation we solve Eq. (21) with m = 0 together with the re-placement equation for the initial-velocity condition, Eq. (18). The equationsare

ui(1) + ui(−1) = 1

2

[uE(0) + uW(0) + uN(0) + uS(0)

], (22)

ui(1) − ui(−1) = 2tgi. (23)

Because g(x, y) = 0 in this instance, we find

ui(1) = 1

4

[uE(0) + uW(0) + uN(0) + uS(0)

]as the starting equation; the right-hand side contains known values of u only.In Fig. 8 are representations of the numerical solution at various time levels.

The simple numerical technique we have developed can be adapted easilyto treat inhomogeneities, boundary conditions involving derivatives of u, ortime-varying boundary conditions. Even nonrectangular regions can be han-dled, provided they fit neatly on a rectangular grid. Several exercises illustratethese points.

E X E R C I S E S

In Exercises 1–5, set up replacement equations using the given space mesh andthe numbering shown in the figure cited. Then find the ui(m) for a few valuesof m using the largest stable value of r. Let boundary conditions override theinitial condition if there is a disagreement.

1. ∇2u = ∂u

∂t, 0 < x < 1, 0 < y < 0.75, 0 < t,

u(0, y, t) = 0, u(1, y, t) = 0, 0 < y < 0.75, 0 < t,

u(x,0, t) = 0, u(x,0.75, t) = 1, 0 < x < 1, 0 < t,

u(x, y,0) = 0, 0 < x < 1, 0 < y < 0.75,

x = y = 1/4. (See Fig. 9a.)

2. ∇2u = ∂u

∂tin R, 0 < t

u = 0 on boundary, 0 < t

u = 1 in R, t = 0.


Figure 8 Displacements of the square membrane. Numbers shown areui(m) × 64.

The region R is an inverted T: Starting with a rectangle of width 1 andheight 3/4, remove a 1/4 × 1/4 square from the upper left and right cor-ners. Take x = y = 1/4. (See Fig. 9b.)


Figure 9 Regions for Exercises 1–3.

3. Same as Exercise 2, except that the region is a cross. (See Fig. 9c.)

4. Same as Eqs. (9)–(11), except that the boundary condition is u = 1 on thebottom (y = 0) and u = 0 elsewhere. (See Fig. 7.)

5. ∇2u = ∂u

∂t, 0 < x < 1, 0 < y < 1, 0 < t,

u(0, y, t) = 0, u(1, y, t) = 1, 0 < y < 1, 0 < t,

u(x,0, t) = 0, u(x,1, t) = 1, 0 < x < 1, 0 < t,

u(x, y,0) = 0, 0 < x < 1, 0 < y < 1,

x = y = 1/4. (See Fig. 2.)

6. Find a numerical solution of the heat problem on a 1×1 square with x =y = 1/4. Initially u = 0 and on the outside boundary u = 0. There is atiny hole in the center of the square, so u(1/2,1/2, t) = 1, t > 0. (Actually,the region is a punctured square.)

7. Solve numerically Eqs. (14)–(18) with x = y = 1/4, ρ2 = 1/2. Takef (x, y) ≡ 0 and

g(x, y) ={

4√

2 at(

12 ,

12

),

0 elsewhere.

Physically, u describes the vibrations of a square membrane struck in themiddle.

8. Obtain an approximate solution of Eqs. (14)–(18) with f (x, y) ≡ 0 andg(x, y) = 4

√2. Take x = y = 1/4 and ρ2 = 1/2.

9. Same as Exercise 8, but f (x, y) ≡ 1 and g(x, y) ≡ 0 in the square.

10. Obtain an approximate numerical solution to the wave equation on an L-shaped region (a 1 × 1 square with a 1/4 × 1/4 square removed from theupper right corner). Assume initial displacement = 1 in the lower rightcorner, initial velocity equal to 0, and zero displacement on the boundary.Take x = y = 1/4 and ρ2 = 1/2.


11. Approximate the solution of the wave equation in a semi-infinite strip 3units wide. Assume u = 0 on all boundaries, zero initial velocity, and aninitial value for u that is 1 in a corner and 0 elsewhere. Take x = y = 1and ρ2 = 1/2.

7.6 Comments and ReferencesOur objective in this chapter has been to survey some elementary numericalmethods for problems like those we attacked analytically in earlier chapters.We have only had enough space to touch on the central topics: obtaining re-placement equations, solving linear systems of equations by direct and iterativemethods, numerical stability, and order of error.

The methods we have introduced are satisfactory for a first introduction andfor learning something about partial differential equations, but they are notadequate for any serious problem solving. New techniques for these problemsare superior in speed, accuracy and stability but are also more complicated. Ofthe many texts available, two excellent ones are Numerical Analysis by Burdenand Faires, for general methods, and Numerical Solution of Partial DifferentialEquations by Smith. (See the Bibliography.)

Almost all numerical methods for linear partial differential equations relyon the symbolism and theory of matrices. Two outstanding texts on matrixtheory are Applied Linear Algebra, 3rd ed., by Noble and Daniel, and Matricesby Barnett.

Miscellaneous Exercises1. Set up and solve replacement equations for this boundary value problem.

Use x = 1/3.

d2u

dx2− √

24x u = 0, 0 < x < 1,

du

dx(0) = 1, u(1) = 1.

2. Use the change of variables x = (r − a)/(b − a) and v(r) = u(x) to con-vert the equation

1

r

d

dr

(r

dv

dr

)− q(r)v = f (r), a < r < b,

to an equation in u on the interval 0 < x < 1.


3. By means of the transformation mentioned in Exercise 2, a heat problemon an annular ring is converted to

d2u

dx2+ 1

1 + x

du

dx= −(1 + x), 0 < x < 1,

u(0) = 1, u(1) = 0.

Set up and solve replacement equations for this problem using x =1/4.

4. The boundary value problem

1

r

d

dr

(r

dv

dr

)− γ 2v = 0, a < r < b,

v(a) = 1, v(b) = 0,

can be transformed into the problem

d2u

dx2+ 1

α + x

du

dx− γ 2L2u = 0, 0 < x < 1,

u(0) = 1, u(1) = 0,

where L = b − a and α = a/L. Set up and solve replacement equationsusing x = 1/4, α = 1, γ L = 1.

5. Set up replacement equations for the heat problem in the following, andsolve for t up to 1/4, using x = 1/4, t = 1/32.

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = u(1, t) = 1 − e−t, 0 < t,

u(x,0) = 0, 0 < x < 1.

6. Same as Exercise 5, but use u(0, t) = u(1, t) = 1 − e−32(ln 2)t so thatu(0, tm) = 1 − (0.5)m.

7. Compare the numerical solution of the problem

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 0, 0 < t,

u(x,0) = 1, 0 < x < 1,


with the solution of the problem consisting of the equation

∂2u

∂x2− 16u = ∂u

∂t, 0 < x < 1, 0 < t,

with the same initial and boundary conditions. Use x = 1/4, t =1/48 in both cases.

8. In Exercise 7, what is the longest stable time step for each of the twoproblems?

9. Solve for several time levels using x = 1/5 and r = 1/2. What is t?

∂2u

∂x2= ∂u

∂t, 0 < x < 1, 0 < t,

u(0, t) = 25t, u(1, t) = 0, 0 < t,

u(x,0) = 0, 0 < x < 1.

10. Same as Exercise 9, except the second boundary condition is ∂u∂x (1, t) = 0.

11. This problem describes the displacement of a string whose end is jerked:

∂2u

∂x2= ∂2u

∂t2, 0 < x < 1, 0 < t,

u(0, t) = 0, u(1, t) = 1, 0 < t,

u(x,0) = 0,∂u

∂t(x,0) = 0, 0 < x < 1.

Solve numerically through one period (until t = 2) with x = t =1/4.

12. Same problems as Exercise 11, except the right-hand boundary condi-tion is u(1, t) = h(t), 0 < t, where

h(t) ={1, 0 < t ≤ 1,

0, 1 < t ≤ 2,

and h(t + 2) = h(t). Solve numerically with x = t = 1/4 for enoughvalues of t so that resonance becomes noticeable.

13. Using x = y = 1/4, find a numerical solution of this problem:

∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < 1, 0 < y < 1,

u(x,0) = 0, u(x,1) = 2

πtan−1

(1

x

), 0 < x < 1,

u(0, y) = 1, u(1, y) = 2

πtan−1(y), 0 < y < 1.


14. The analytical solution of the problem in Exercise 13 is u(x, y) =(2/π) tan−1(y/x). Compare your numerical results with the exact so-lution.

15. Using x = y = 1/4 and r = 1/4, find a numerical solution for thisproblem:

∇2u = ∂u

∂t, 0 < x < 1, 0 < y < 1, 0 < t,

u(x,0, t) = u(x,1, t) = 0, 0 < x < 1, 0 < t,

u(0, y, t) = u(1, y, t) = 0, 0 < y < 1, 0 < t,

u(x, y,0) = 1, 0 < x < 1, 0 < y < 1.

16. The analytical solution of the problem in Exercise 15 is

u(x, y, t) =∞∑

n=1

∞∑n=1

4(1 − cos(nπ))(1 − cos(mπ))

π2mn

× sin(nπx) sin(nπy)e−(m2+n2)π2t .

Using just the term m = n = 1 of this solution, compare the ratio

R = u(

12 ,

12 , tm+1

)u(

12 ,

12 , tm

)and the ratio of the corresponding u’s computed in Exercise 15.

Bibliography

Abramowitz, M., and I. Stegun (eds). Handbook of Mathematical Functions,10th ed. Washington, DC, National Bureau of Standards, 1972. (Reprintedby Dover, 1974.)

Andrews, L.C. Special Functions of Mathematics for Engineers, 2nd ed. Belling-ham, WA, SPIE — The International Society for Optical Engineering, 1997.

Barnett, S. Matrices: Methods and Applications. New York, Oxford UniversityPress, 1990.

Burden, R.L., and J.D. Faires. Numerical Analysis, 7th ed. Belmont, CA,Brooks/Cole, 2000.

Carslaw, H.S., and J.C. Jaeger. Conduction of Heat in Solids, 2nd ed. New York,Oxford University Press, 1986.

Churchill, R.V., and J.W. Brown. Fourier Series and Boundary Value Problems,6th ed. New York, McGraw-Hill, 2000.

Churchill, R.V. Operational Mathematics, 3rd ed. New York, McGraw-Hill,1972.

Courant, R., and D. Hilbert. Methods of Mathematical Physics, Vol. 1. NewYork, Wiley-Interscience, 1953/1989.

Crank, J. The Mathematics of Diffusion, 2nd ed. New York, Oxford UniversityPress, 1980.

Davis, P.J., and R. Hersh. The Mathematical Experience. Boston, HoughtonMifflin, 1999.

Erdelyi, A., W. Magnus, F. Oberhettinger, and F. Tricomi. Tables of IntegralTransforms, Vols. 1 and 2. New York, McGraw-Hill, 1954.

Feller, W. Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed.New York, Wiley, 1968.

433

434 Bibliography

Fletcher, N.H., and T.D. Rossing. The Physics of Musical Instruments, 2nd ed.New York, Springer-Verlag, 1998.

Isenberg, C. The Science of Soap Films and Soap Bubbles. Avon, UK, Tieto Ltd,1978. (Reprinted by Dover, 1992.)

Jerri, A.J. Integral and Discrete Transforms with Applications and Error Analysis.New York, Marcel Dekker, 1991.

Jones, D.S., and B.D. Sleeman. Differential Equations and Mathematical Biol-ogy. San Francisco, Harper-Collins, 1983.

Kirchhoff, R.H. Potential Flows: Computer Graphic Solutions. New York, MarcelDekker, 1998.

Lamb, H. Hydrodynamics, 6th ed. Cambridge, UK, Cambridge UniversityPress, 1971.

Love, A.E.H. A Treatise on the Mathematical Theory of Elasticity, 4th ed. NewYork, Dover, 1944.

Main, I.G. Vibrations and Waves in Physics, 3rd ed. Cambridge, UK, CambridgeUniversity Press, 1993.

Morse, P.M., and H. Feshbach. Methods of Theoretical Physics, Part I. New York,McGraw-Hill, 1953.

Murray, J.D. Mathematical Biology. New York, Springer-Verlag, 1993.Noble, B., and J.W. Daniel. Applied Linear Algebra, 3rd ed. Englewood Cliffs,

NJ, Prentice-Hall, 1988.Sagan, H. Boundary and Eigenvalue Problems in Mathematical Physics. New

York, Wiley, 1966. (Reprinted by Dover, 1989.)Smith, G.D. Numerical Solution of Partial Differential Equations, 3rd ed. New

York, Oxford University Press, 1986.Street, R.L. Analysis and Solution of Partial Differential Equations. Monterey,

CA, Brooks/Cole, 1973.Timoshenko, S., and J.N. Goodier. Theory of Elasticity, 2nd ed. New York,

McGraw-Hill, 1951.Tolstov, G.P. Fourier Series. Englewood Cliffs, NJ, Prentice-Hall, 1962.

(Reprinted by Dover, 1976.)Walker, J.S. The Fast Fourier Transform, 2nd ed. Boca Raton, FL, CRC Press,

1996.Wan, F.Y.M. Mathematical Models and Their Analysis. New York, Harper &

Row, 1989.Widder, D.V. The Heat Equation. New York, Academic Press, 1975.

Appendix:MathematicalReferences

Trigonometric Functions

sin(A ± B) = sin(A) cos(B) ± cos(A) sin(B)

cos(A ± B) = cos(A) cos(B) ∓ sin(A) sin(B)

sin(A) + sin(B) = 2 sin

(A + B

2

)cos

(A − B

2

)

sin(A) − sin(B) = 2 cos

(A + B

2

)sin

(A − B

2

)

cos(A) + cos(B) = 2 cos

(A + B

2

)cos

(A − B

2

)

cos(A) − cos(B) = 2 sin

(A + B

2

)sin

(A − B

2

)

sin(A) sin(B) = 1

2

(cos(A − B) − cos(A + B)

)sin(A) cos(B) = 1

2

(sin(A − B) + sin(A + B)

)cos(A) cos(B) = 1

2

(cos(A − B) + cos(A + B)

)cos(A) = 1

2

(eiA + e−iA

), sin(A) = 1

2i

(eiA − e−iA

)cos2(A) + sin2(A) = 1, 1 + tan2(A) = sec2(A)

435

436 Appendix: Mathematical References

Hyperbolic Functions

cosh(A) = 1

2

(eA + e−A

), sinh(A) = 1

2

(e A − e−A

)d cosh(u) = sinh(u)du, d sinh(u) = cosh(u)du

sinh(A ± B) = sinh(A) cosh(B) ± cosh(A) sinh(B)

cosh(A ± B) = cosh(A) cosh(B) ± sinh(A) sinh(B)

sinh(A) + sinh(B) = 2 sinh

(A + B

2

)cosh

(A − B

2

)

sinh(A) − sinh(B) = 2 cosh

(A + B

2

)sinh

(A − B

2

)

cosh(A) + cosh(B) = 2 cosh

(A + B

2

)cosh

(A − B

2

)

cosh(A) − cosh(B) = 2 sinh

(A + B

2

)sinh

(A − B

2

)

sinh(A) sinh(B) = 1

2

(cosh(A + B) − cosh(A − B)

)sinh(A) cosh(B) = 1

2

(sinh(A + B) + sinh(A − B)

)cosh(A) cosh(B) = 1

2

(sinh(A + B) + cosh(A − B)

)cosh2

(A) − sinh2(A) = 1, 1 − tanh2

(A) = sech2(A)

Calculus1. Derivative of a product

(uv)′ = u′v + uv′

(uv)′′ = u′′v + 2u′v′ + uv′′

(uv)(n) = u(n)v +(

n

1

)u(n−1)v′ + · · · +

(n

n − 1

)uv(n−1) + uv(n)

In this formula,(n

k

) = n!(n−k)!k! is a binomial coefficient.

2. Rules of integration

a.

∫ b

a

(c1f1(x) + c2f2(x)

)dx = c1

∫ b

af1(x)dx + c2

∫ b

af2(x)dx


b.

∫ a

af (x)dx = 0

c.

∫ b

af (x)dx = −

∫ a

bf (x)dx

d.

∫ b

af (x)dx =

∫ c

af (x)dx +

∫ b

cf (x)dx

3. Derivatives of integrals

a.d

dt

∫ b

af (x, t)dx =

∫ b

a

∂f

∂t(x, t)dx

b.d

dt

∫ t

af (x)dx = f (t) (Fundamental theorem of calculus;

a is constant)

c.d

dt

∫ v(t)

u(t)f (x, t)dx = f

(v(t), t

)v′(t) − f

(u(t), t

)u′(t)

+∫ v(t)

u(t)

∂f

∂t(x, t)dx (Leibniz’s rule)

4. Integration by parts

a.

∫uv′ dx = uv −

∫vu′ dx

b.

∫uv′′ dx = v′u − vu′ +

∫vu′′ dx

5. Functions defined by integrals

a. Natural logarithm

ln(x) =∫ x

1

dz

z

b. Sine-integral function

Si(x) =∫ x

0

sin(z)

zdz

c. Normal probability distribution function

�(x) = 1√2π

∫ x

−∞e−z2/2 dz

d. Error function

erf(x) = 2√π

∫ x

0e−z2

dz

Note: erf(x) = 2�(√

2x) − 1


e. Integrated Bessel function

IJ(x) =∫ x

0J0(z)dz

Table of IntegralsAny letter except x represents a constant. The integration constants have beenleft off.

1. Rational functions

1.1

∫dx

h + kx= 1

kln |h + kx|

1.2

∫dx

x2 + a2= 1

atan−1

(x

a

)

1.3

∫x dx

x2 + a2= 1

2ln

(x2 + a2

)

1.4

∫dx

x2 − a2= 1

2aln

∣∣∣∣x − a

x + a

∣∣∣∣1.5

∫x dx

x2 − a2= 1

2ln

∣∣x2 − a2∣∣

2. Radicals

2.1

∫dx√

x2 + a2= ln

(x +

√x2 + a2

)or sinh−1

(x

a

)

2.2

∫x dx√x2 + a2

=√

x2 + a2

2.3

∫dx√

x2 − a2= ln

(x +

√x2 − a2

)(x > a)

2.4

∫x dx√x2 − a2

=√

x2 − a2

2.5

∫dx√

a2 − x2= sin−1

(x

a

)(|x| < a)

2.6

∫x dx√a2 − x2

= −√

a2 − x2 (|x| < a)

3. Exponentials and hyperbolic functions

3.1

∫ekx dx = ekx

k


3.2

∫xekx dx = kx − 1

k2ekx

3.3

∫sinh(kx)dx = cosh(kx)

k

3.4

∫cosh(kx)dx = sinh(kx)

k

3.5

∫x sinh(kx)dx = x cosh(kx)

k− sinh(kx)

k2

3.6

∫x cosh(kx)dx = x sinh(kx)

k− cosh(kx)

k2

4. Sines and cosines

4.1

∫sin(λx)dx = − cos(λx)

λ

4.2

∫cos(λx)dx = sin(λx)

λ

4.3

∫x sin(λx)dx = sin(λx)

λ2− x cos(λx)

λ

4.4

∫x cos(λx)dx = cos(λx)

λ2+ x sin(λx)

λ

4.5

∫x2 sin(λx)dx = 2x sin(λx)

λ2+ (2 − λ2x2) cos(λx)

λ3

4.6

∫x2 cos(λx)dx = 2x cos(λx)

λ2+ (λ2x2 − 2) sin(λx)

λ3

4.7

∫sin(λx) sin(µx)dx = sin(µ − λ)x

2(µ − λ)− sin(µ + λ)x

2(µ + λ)(λ �= µ)

4.8

∫sin(λx) cos(µx)dx = cos(µ − λ)x

2(µ − λ)− cos(µ + λ)x

2(µ + λ)(λ �= µ)

4.9

∫cos(λx) cos(µx)dx = sin(µ − λ)x

2(µ − λ)+ sin(µ + λ)x

2(µ + λ)(λ �= µ)

4.10

∫sin2(λx)dx = x

2− sin(2λx)

4λ

4.11

∫sin(λx) cos(λx)dx = sin2(λx)

2λ

4.12

∫cos2(λx)dx = x

2+ sin(2λx)

4λ


4.13

∫ekx sin(λx)dx = ekx(k sin(λx) − λ cos(λx))

k2 + λ2

4.14

∫ekx cos(λx)dx = ekx(k cos(λx) + λ sin(λx))

k2 + λ2

4.15

∫sinh(kx) sin(λx)dx = k cosh(kx) sin(λx) − λ sinh(kx) cos(λx)

k2 + λ2

4.16

∫sinh(kx) cos(λx)dx = k cosh(kx) cos(λx) + λ sinh(kx) sin(λx)

k2 + λ2

4.17

∫cosh(kx) sin(λx)dx = k sinh(kx) sin(λx) − λ cosh(kx) cos(λx)

k2 + λ2

4.18

∫cosh(kx) cos(λx)dx = k sinh(kx) cos(λx) + λ cosh(kx) sin(λx)

k2 + λ2

5. Bessel functions

5.1

∫xJ0(λx)dx = xJ1(λx)

λ

5.2

∫x2J0(λx)dx = x2J1(λx)

λ+ xJ0(λx)

λ2− 1

λ3IJ(λx)1

5.3

∫J1(λx)dx = − J0(λx)

λ

5.4

∫xn+1Jn(λx)dx = xn+1Jn+1(λx)

λ

5.5

∫Jn(λx)

dx

xn−1= − Jn−1(λx)

λxn−1

5.6

∫J2

0(λx)x dx = x2

2

[J2

0(λx) + J21(λx)

]5.7

∫J2

n(λx)x dx = x2

2

[J2

n(λx) − Jn−1(λx)Jn+1(λx)]

= x2

2

[J ′

n(λx)]2 +

(x2

2− n2

2λ2

)[Jn(λx)

]2

6. Legendre polynomials

6.1

∫Pn(x)dx = −−(1 − x2)

n(n + 1)P′

n(x)

6.2

∫xPn(x)dx = (1 − x2)

(n + 2)(n − 1)

(Pn(x) − xP′

n(x)).

1See Calculus 5e.

Answers toOdd-NumberedExercises

Chapter 0

Section 0.11. φ(x) = c1 cos(λx) + c2 sin(λx).

3. The equation has constant coefficients k = 0, p = 0; u(t) = c1 + c2t.

5. w(r) = c1rλ + c2r−λ.

7. Integrate, solve for dv/dx, and integrate again:

v(x) = c1 + c2 ln |h + kx|.9. u(x) = c1 + c2/x2.

11. u(r) = c1 + c2 ln(r).

13. Characteristic polynomial m4 − λ4 = 0; roots m = ±λ,±iλ. General so-lution u(x) = c1 cos(λx) + c2 sin(λx) + c3 cosh(λx) + c4 sinh(λx).

15. Characteristic polynomial (m2 + λ2)2 = 0; roots m = ±iλ (double).General solution u(x) = (c1 + c2x) cos(λx) + (c3 + c4x) sin(λx).

17. v(t) = ln(t) and u2(t) = tb ln(t).

19. u′′ + λ2u = 0; R(ρ) = (a cos(λρ) + b sin(λρ))/ρ.

21. t2d2u/dt2 = v′′ − v′; t du/dt = v′; v′′ + (k − 1)v′ + pv = 0 (constantcoefficients).

441

442 Answers to Odd-Numbered Exercises

23. Roots of characteristic equation:

m = −α ± iβ, β =√

σ 2 − α2.

Solution of differential equation:

y(t) = e−αt(c1 cos(βt) + c2 sin(βt)

).

Initial conditions give: c1 = −0.001h, c2 = (α/β)c1.

25. v = 2.62 m/s.

Section 0.21. u(t) = T + ce−at .

3. u(t) = te−at + ce−at .

5. u(t) = 1

2t sin(t) + c1 cos(t) + c2 sin(t).

7. u(t) = 1

12et + 1

2te−t + c1e−t + c2e−2t .

9. u(ρ) = −1

6ρ2 + c1

ρ+ c2.

11. h(t) = −320t + c1 + c2e−0.1t , c1 = h0 + 3200, c2 = −3200.

13. v(t) = t, up(t) = te−at .

15. v1(x) = sin(x) − ln |sec(x) + tan(x)|, v2 = − cos(x);

up(x) = − cos(x) ln |sec(x) + tan(x)|.17. v1(t) = t2/2, v2(t) = −t; up(t) = −t2/2.

19. v1(t) = −1/2t, v2(t) = −t/2, up(t) = −1.

23. β = 1/α, K = Rα/ρc.

25. T = β(exp(KI2max(1 − e−2λt)/2λ) − 1).

Section 0.31. a. u(x) = c2 sin(x), c2 arbitrary;

b. u(x) = 1 − cos(x) − 1 − cos(1)

sin(1)sin(x) (unique);

c. No solution exists.

3. a. and b. λ = ±(2n − 1)π

2a, n = 1,2, . . . ;


c. λ = ±nπ

a, n = 0,1,2, . . . .

5. c = −a/2, c′ = h − 1

µcosh

(µa

2

).

7. u(x) = T + c1 cosh(γ x) + c2 sinh(γ x), where

γ =√

hC

κAand c1 = T0 − T, c2 = −κγ sinh(γ a) + h cosh(γ a)

κγ cosh(γ a) + h sinh(γ a)c1.

9. u(x) = T + H

(1 − cosh(γ x) − 1 − cosh(γ a)

sinh(γ a)sinh(γ x)

),

where H = I2R

hCand γ =

√hC

κA.

11. u(y) = y(L − y)g/2µ.

13. P = EI(nπ/L)2, n = 1,2, . . . .

15. u(x) = T + A

(1 − cosh(γ x) − 1 − cosh(γ a)

sinh(γ a)sinh(γ x)

),

A = g/κγ 2, and γ =√

hC

κA.

17. u(r) = c1 ln(r/a) + c2, c1 = h0h1(Ta − TW)/D, c2 = [h0(κ/b +h1 ln(b/a))TW + (κ/a)h1Ta]/D, D = h1κ/a + h0κ/b + h0h1 ln(b/a).

19. u(x) = w0

EI

(x4

24− ax3

6+ a2x2

2

).

Section 0.41. a. u′′ + 1

ru′ − u = 0, r = 0;

b. u′′ − 2x

1 − x2u′ = 0, x = ±1;

c. u′′ + cot(φ)u′ − u = 0, φ = 0, ±π,±2π, . . . ;

d. u′′ + 2

ρu′ + λ2u = 0, ρ = 0.

3. u(0) bounded; u(ρ) = H

6κ

(c2 − ρ2

) + Hc

3h+ T.

5. u(ρ) = 1

ρ

(c1 cos(µρ) + c2 sin(µρ)

),

u(ρ) ≡ 0 unless µa = π,2π, . . . . The critical radius is a = π

µ.


7. u(r) = 325 + 104(0.25 − r2)/4; u(0) = 950.

9. u(x) = T0 + AL2(1 − e−x/L).

Section 0.51. G(x, z) =

{z(a − x)/(−a), 0 < z ≤ x,x(a − z)/(−a), x ≤ z < a.

3. G(x, z) ={

cosh(γ z) sinh(γ (a − x))/(−γ cosh(γ a)), 0 < z ≤ x,cosh(γ x) sinh(γ (a − z))/(−γ cosh(γ a)), x ≤ z < a.

5. G(ρ, z) =

(c − ρ)/ρ

−c/z2, 0 ≤ z < ρ,

(c − z)/z

−c/z2, ρ ≤ z < c.

7. G(x, z) =

sinh(γ z)e−γ x

−γ, 0 < z ≤ x,

sinh(γ x)e−γ z

−γ, x ≤ z.

9. u(ρ) = (ρ2 − c2)/6.

11. u(x) =∫ a

0G(x, z)f (z)dz =

∫ x

0

z(a − x)

−af (z)dz +

∫ a

x

x(a − z)

−af (z)dz.

There are two cases:

(i) x ≤ a/2, so u(x) =∫ a

a/2

x(a − z)

−adz;

and

(ii) x > a/2, so u(x) =∫ a

x

x(a − z)

−adz.

Results: u(x) ={−ax/8, 0 < x < a/2,

−x(a − x)2/2a, a/2 < x < a.

13. (i) At the left boundary, x = l < z, so the second line of Eq. (17) holds.The boundary condition (2) is satisfied by v because it is satisfied by u1.At the right boundary, use the first line of Eq. (17).

(ii) At x = z, both lines of Eq. (17) give the same value.

(iii) v′(z + h) − v′(z − h) = u1(z)u′2(z + h) − u′

1(z − h)u2(z)

W(z).

As h approaches 0, the numerator approaches W(z).


(iv) This is true because u1(x) and u2(x) are solutions of the homoge-neous equation.

Chapter 0 Miscellaneous Exercises

1. u(x) = T0 cosh(γ x) + (T1 − T0 cosh(γ a)

) sinh(γ x)

sinh(γ a).

3. u(x) = T0.

5. u(r) = p(a2 − r2)/4.

7. u(ρ) = H(a2 − ρ2)/6 + T0.

9. u(x) = T + (T1 − T) cosh(γ x)/ cosh(γ a).

11. u(x) = T0 + (T − T0)e−γ x.

13. h(x) = √ex(a − x) + h2

0 + (h21 − h2

0)(x/a).

15. u(x) = w(1 − e−γ x cos(γ x))EI/k, where γ = (k/4EI)1/4.

17. u(x) ={

T0 + Ax, 0 < x < αa,T1 − B(a − x), αa < x < a,

A = κ2

κ1(1 − α) + κ2α

T1 − T0

a, B = κ1

κ2A.

19. u(x) = 1

2

(1 − e−2x

) − 1

2

(1 − e−2a

)1 − e−x

1 − e−a.

21. a. u(x) = sinh(px)/ sinh(pa);

b. u(x) = cosh(px) − cosh(pa)

sinh(pa)sinh(px) = sinh

(p(a − x)

)/ sinh(pa);

c. u(x) = cosh(px)/ cosh(pa);

d. u(x) = cosh(p(a − x))/ cosh(pa);

e. u(x) = − cosh(p(a − x))/p sinh(pa);

f. u(x) = cosh(px)/p sinh(pa).

23. u(x) = x

2ln

∣∣∣∣1 + x

1 − x

∣∣∣∣ − 1.

25. Multiply by u′ and integrate: 12 (u′)2 = 1

5γ2u5 + c1. Since u(x) → 0

as x → ∞, also u′(x) → 0; thus c1 = 0. Now u′ = −√2γ 2/5u5/2 or

u−5/2u′ = −√2γ 2/5 (the negative root makes u decrease) can be in-

tegrated to result in (−2/3)u−3/2 = −√2γ 2/5x + c2. The condition at

x = 0 gives c2 = (−3/2)U−3/2.


Finally u(x) = (U−3/2 + (3/2)√

2γ 2/5x)−2/3.

27. 459.77 rad/s.

29. u(x) = C0e−ax.

31. w(x) = P

2γ 2

[1

4− x2 + cosh(γ x) − cosh(γ/2)

γ sinh(γ/2)

].

33. The solution breaks down (buckling occurs) if tan(λ/2) = γ/2.

Chapter 1

Section 1.11. a. 2

(sin(x) − 1

2sin(2x) + 1

3sin(3x) − · · ·

);

b.π

2− 4

π

(cos(x) + 1

9cos(3x) + 1

25cos(5x) + · · ·

);

c.1

2+ 2

π

(sin(x) + 1

3sin(3x) + 1

5sin(5x) + · · ·

);

d.2

π− 4

π

(1

3cos(2x) + 1

15cos(4x) + 1

35cos(6x) + · · ·

).

3. f (x + p) = 1 = f (x) for any p and all x.

5. If c is a multiple of p, the graph of f (x) between c and c + p is the sameas that between 0 and p. Otherwise, let k be the integer such that kp liesbetween c and c + p:∫ c+p

cf (x)dx =

∫ kp

cf (x)dx +

∫ c+p

kpf (x)dx =

∫ p

c∗f (x)dx +

∫ c∗

0f (x)dx,

where c∗ = c − (k − 1)p.

7. a. cos2(x) = 1

2+ 1

2cos(2x);

b. sin

(x − π

6

)= cos

(π

6

)sin(x) − sin

(π

6

)cos(x);

c. sin(x) cos(2x) = −1

2sin(x) + 1

2sin(3x).

Section 1.21. a.

1

2− 4

π2

[cos(πx) + 1

9cos(3πx) + 1

25cos(5πx) + · · ·

];

Chapter 1 447

b.4

π

[sin

(πx

2

)+ 1

3sin

(3πx

2

)+ 1

5sin

(5πx

2

)+ · · ·

];

c.1

12− 1

π2

[cos(2πx) − 1

4cos(4πx) + 1

9cos(6πx) − · · ·

].

3. f (x) = f (x − 2na), 2na < x < 2(n + 1)a,

f (x) ∼ a0 +∞∑1

an cos(nπx/a) + bn sin(nπx/a),

a0 = 1

2a

∫ 2a

0f (x)dx, an = 1

a

∫ 2a

0f (x) cos(nπx/a)dx,

bn = 1

a

∫ 2a

0f (x) sin(nπx/a)dx.

5. Odd: (a), (d), (e); even: (b), (c); neither: (f).

7. a.2

π

(sin(πx) − 1

2sin(2πx) + · · ·

);

b. This function is its own Fourier series;

c.4

π2

(sin(πx) − 1

9sin(3πx) + 1

25sin(5πx) − · · ·

).

9. If f (−x) = −f (x) and f (x) = f (a − x) for 0 < x < a, sine coefficientswith even indices are zero. Example: square wave.

11. a. f (x) = 1 = 2

π

∞∑1

1 − cos(nπ)

nsin

(nπx

a

);

b. f (x) = a

2− 2a

π2

∞∑1

1 − cos(nπ)

n2cos

(nπx

a

)

= 2a

π

∞∑1

− cos(nπ)

nsin

(nπx

a

);

c. f (x) =∞∑1

(−1)n+1 sin(1)2nπ

(nπ)2 − 1sin(nπx), 0 < x < 1

=∞∑1

((−1)n cos(1) − 1

) 2

(nπ)2 − 1cos(nπx), 0 < x < 1;

d. f (x) = 2

π

[1 −

∞∑1

1 + cos(nπ)

n2 − 1cos(nx)

]= sin(x).

13. Even, yes. Odd, yes only if f (0) = f (a) = 0.


Section 1.31. a. sectionally smooth; b, c, d, e are not;

b: vertical tangent at 0; c: vertical asymptote at ±π/2; d, e: vertical asymp-tote at π/2.

3. To f (x) everywhere.

5. b. Graph consists of straight-line segments. c. x = 1, sum = 1/2; x = 2,sum = 0; x = 9.6, sum = −0.6; x = −3.8, sum = 0.2. Use periodicity.

7. B = 0, A = −π2/12, C = 1/4.

9. a.√

1 − x2; b. a0 = π/4; c. No; d. nothing.

Section 1.41. (c), (d), (f), (g) have uniformly convergent Fourier series.

3. All of the cosine series converge uniformly. The sine series converges uni-formly only in case (b).

5. (a), (b), (d) converge uniformly; (c) does not.

Section 1.5

1.∞∑

n=1

1

n2= π2

6.

3. f ′(x) = 1, 0 < x < π . The sine series cannot be differentiated, because theodd periodic extension of f is not continuous. But the cosine series can bedifferentiated.

5. For the sine series: f (0+) = 0 and f (a−) = 0. For the cosine series noadditional condition is necessary.

7. No. The function ln |2 cos( x2 )| is not even sectionally continuous.

9. Since f is odd, periodic, and sectionally smooth, (c) follows, and alsobn → 0 as n → ∞. Then

∑∞n=1 |nkbne−n2t| converges for all integers k

(t > 0) by the comparison test and ratio test:∣∣nkbne−n2t∣∣ ≤ Mnke−n2t for some M

and

M(n + 1)ke−(n+1)2t

Mnke−n2t=

(n + 1

n

)k

e−(2n+1)t → 0

as n → ∞. Then by Theorem 7, (a) is valid. Property (b) follows by direc-tion substitution.

Chapter 1 449

Section 1.6

1.1

π

∫ π

−π

(ln

∣∣∣∣2 cos

(x

2

)∣∣∣∣)2

dx =∞∑

n=1

1

n2= π2

6.

3. a. Coefficients tend to zero.

b. Coefficients tend to zero, although

∫ 1

−1|x|−1dx is infinite.

5. The integral must be infinite, because∞∑

n=1

a2n + b2

n = ∞.

Section 1.71. The equality to be proved is

2 sin

(1

2y

)(1

2+

N∑n=1

cos(ny)

)= sin

((N + 1

2

)y

).

The left-hand side is transformed as follows:

2 sin

(1

2y

)(1

2+

N∑n=1

cos(ny)

)

= sin

(1

2y

)+

N∑n=1

2 sin

(1

2y

)cos(ny)

= sin

(1

2y

)+

N∑n=1

(sin

((n + 1

2

)y

)− sin

((n − 1

2

)y

))

= sin

(1

2y

)+

N∑n=1

sin

((n + 1

2

)y

)−

N−1∑n=0

sin

((n + 1

2

)y

)

= sin

((N + 1

2

)y

)

because all other terms cancel.

3. φ(0+) = 1, φ(0−) = −1. See Fig. 1.

5. a. f ′(x) = 34 x−1/4 for 0 < x < π (and f ′ is an odd function). Thus, f has a

vertical tangent at x = 0, although it is continuous there.

b. φ(y) = |y|3/4

2 sin( 12 y)

cos

(1

2y

), −π < y < π


Figure 1 Graph for Exercise 3, Section 1.7.

is a product of continuous functions and is therefore continuous, exceptperhaps where the denominator is 0. At y = 0, cos( 1

2 y) ∼= 1, 2 sin( 12 y) ∼= y,

so φ(y) ∼= |y|3/4/y = ±|y|−1/4 near y = 0.

c. Now,∫ π

−πφ2(y)dy is finite, so the Fourier coefficients of φ approach

zero.

Section 1.81. a6 = −0.00701, a6 = −0.00569.

3. a0 = 1.367,

a1 = −0.844, b1 = −0.043,

a2 = 0.208, b2 = −0.115,

a3 = 0.050, b3 = −0.050,

a4 = 0.042, b4 = 0.00,

a5 = −0.0064, b5 = 0.043,

a6 = 0.0167.

Section 1.91. Each function has the representations (for x > 0)

f (x) =∫ ∞

0A(λ) cos(λx)dλ =

∫ ∞

0B(λ) sin(λx)dλ.

a. A(λ) = 2/π(1 + λ2), B(λ) = 2λ/π(1 + λ2);

b. A(λ) = 2 sin(λ)/πλ, B(λ) = 2(1 − cos(λ))/πλ;

c. A(λ) = 2(1 − cos(λπ))/λ2π , B(λ) = 2(πλ − sin(λπ))/πλ2.

3. a.1

1 + x2=

∫ ∞

0e−λ cos(λx)dλ;

Chapter 1 451

b.sin(x)

x=

∫ ∞

0A(λ) cos(λx)dλ, where A(λ) =

{1, 0 < x < 1,0, 1 < x.

5. a. A(λ) ≡ 0, B(λ) = 2 sin(λπ)

π(1 − λ2);

b. A(λ) = 1 + cos(λπ)

π(1 − λ2), B(λ) = sin(λπ)

π(1 − λ2);

c. A(λ) = 2(1 + cos(λπ))

π(1 − λ2), B(λ) ≡ 0.

7. Change variable from x to λ with x = λz.

Section 1.10

1. eαx = 2sinh(απ)

π

(1

2α+

∞∑n=1

(−1)n

α2 + n2

(α cos(nx) − n sin(nx)

)).

3. f (x) =∫ ∞

−∞C(λ)eiλxdλ.

a. C(λ) = 1

2π(1 + iλ); b. C(λ) = 1 + e−iλπ

2π(1 − λ2).

5. a. 1 +∞∑

n=1

rn cos(nx) = Re∞∑0

(reix

)n = Re1

1 − reix;

b.∞∑

n=1

sin(nx)

n!= Im

∞∑n=1

einx

n!= Im exp

(eix

).

7. a. f (x) = 2 sin(x)

x; b. f (x) = 2

1 + x2.

Section 1.11

1. u(t) = A0 +∞∑

n=1

An cos(nt/2) + Bn sin(nt/2),

A0 = 1

2.08, An = 0.4/π

(1.04 − n2)2 + (0.4n)2,

Bn = − 1

nπ

1.04 − n2

(1.04 − n2)2 + (0.4n)2.

3. u(x) =∞∑

n=1

Bn sin(nπx/L), Bn = 8K sin(nπ/2)

((nπ/L)2 + γ 2)n2π2,

K = w/EI, γ 2 = T/EI.



1. f (x) =∞∑

n=1

bn sin(nx),

bn ={0, n even,

4 sin(nα)

παn2, n odd.

3. Yes. As α → 0, sin(nα)/nα → 1.

5. f (x) =∞∑

n=1

bn sin(nπx/a),

bn = 2h

π2

sin(nπα)

n2

(1

α+ 1

1 − α

).

7. a. bn = 0, an = 0, a0 = 1;

b.∞∑

n=1

bn sin(nπx/a), bn = 2(1 − cos(nπ))

nπ;

c. and d. same as a;

e. same as b;

f. a0 +∞∑

n=1


a0 = 1

2, an = 0, bn = 1 − cos(nπ)

nπ.

9. f (x) = a0 +∞∑

n=1


a0 = 1

2a, an = −2a(1 − cos(nπ))

n2π2, bn = −2a cos(nπ)

nπ,

x = −a, −a/2, 0, a, 2a,sum = a, 0, 0, a, 0.

11. f (x) = a0 +∞∑

n=1

an cos(nx),

a0 = 3

4, an = sin(nπ/2)

nπ,

x = 0, π/2, π, 3π/2, 2π,

sum = 1,3

4,

1

2,

3

4, 1.

Chapter 1 453

13. f (x) =∞∑

n=1

bn sin(nπx), bn = 2(1 + cos(nπ)

)/nπ .

15. f (x) =∞∑

n=1

bn sin(nx),

b2 = 1

2, other bn = 4 sin(nπ/2)

π(4 − n2).

17.N∑1

cos(nx) = ReN∑1

einx = Reeix − eiNx

1 − eix= Re

eix/2 − ei(2N−1)x/2

e−ix/2 − eix/2.

The denominator is now −2i sin(x/2).

19. f (x) =∞∑

n=1

bn sin(nx), bn = 2a sin(na + π)

n2a2 − π2.

21. f (x) =∫ ∞

0

(sin(λa)

λπcos(λx) + 1 − cos(λa)

λπsin(λx)

)dλ.

23. f (x) =∫ ∞

0

2 sin(λπ)

π(1 − λ2)sin(λx)dλ (x > 0).

29. Use

∫ ∞

0

sin(λt)

λdλ = π

2.

31. These answers are not unique.

a.∞∑

n=1

bn sin(nπx), bn = 2/nπ ;

b. a0 +∞∑

n=1

an cos(nπx), a0 = 1

2, an = 2

(1 − cos(nπ)

)/n2π2;

c.

∫ ∞

0B(λ) sin(λx)dλ, B(λ) = 2

(λ − sin(λ)

)/(πλ2

);

d.

∫ ∞

0A(λ) cos(λx)dλ, A(λ) = 2

(1 − cos(λ)

)/(πλ2

).

The integrals of parts c. and d. converge to 0 for x > 1.

33. Use s = 6 in Eq. (7) of Section 8.

a0 = 0.78424, a4 = −0.00924,

a1 = 0.22846, a5 = 0.00744,

a2 = −0.02153, a6 = −0.00347,

a3 = 0.01410.


35. a0 = a

6, an = 2a

n2π2

(cos

(2nπ

3

)− cos

(nπ

3

)).

37. a0 = 5

8, an = 2

n2π2

(3 cos

(nπ

2

)− 2 − cos(nπ)

).

39. a0 = 1

2, an = 2

n2π2

(1 − cos(nπ)

).

41. a0 = a2

6, an = −2a2

n2π2

(1 + cos(nπ)

).

43. a0 = 1

2, an = −1

nπ2 sin

(nπ

2

).

45. bn = 1 + cos(nπ/2) − 2 cos(nπ)

nπ.

47. bn = a

(2 sin(nπ/2)

n2π2− cos(nπ)

nπ

).

49. bn = 2

nπ

(cos

(nπ

4

)− cos

(3nπ

4

)).

51. bn = 2nπ(1 − eka cos(nπ))

(a2k2 + n2π2).

53. A(λ) = 2

π(1 + λ2).

55. A(λ) = 2 sin(λb)

πλ.

57. A(λ) = 2(1 − cos(λ))

πλ2.

59. B(λ) = 2λ

π(1 + λ2).

61. B(λ) = 2(1 − cos(λb))

λπ.

63. B(λ) = 2(λ − sin(λ))

λ2π.

65. The term an cos(nx) + bn sin(nx) appears in Sn,Sn+1, . . . ,SN , and thusN + 1 − n times in σN .

67. Use Eq. (13) of Section 7 and the identity in Exercise 66.

69. a. Use x = 0; b. x = 1/2; c. x = 0.

Chapter 2 455

Chapter 2

Section 2.11. One possibility: u(x, t) is the temperature in a rod of length a whose lat-

eral surface is insulated. The temperature at the left end is held constantat T0. The right end is exposed to a medium at temperature T1. Initiallythe temperature is f (x).

3. Ax g = hCx(U −u(x, t)), where h is a constant of proportionality andC is the circumference. Eq. (4) becomes

∂2u

∂x2+ hC

κA(U − u) = 1

k

∂u

∂t.

5. If ∂x∂u (0, t) is positive, then heat is flowing to the left, so u(0, t) is greater

than T(t).

7. The second factor is approximately constant if T is much larger than u orif T and u are approximately equal.

Section 2.21. v′′ − γ 2(ν − U) = 0, 0 < x < a,

v(0) = T0, v(a) = T1,

v(x) = U + A cosh(γ x) + B sinh(γ x),

A = T0 − U , B = (T1 − U) − (T0 − U) cosh(γ a)

sinh(γ a).

One interpretation: u is the temperature in a rod, with convective heattransfer from the cylindrical surface to a medium at temperature U .

3. v(x) = T. Heat is being generated at a rate proportional to u − T. If γ =π/a, the steady-state problem does not have a unique solution.

5. v(x) = A ln(κ0 + βx) + B, A = (T1 − T0)/ ln(1 + aβ/κ0),

B = T0 − A ln(κ0).

7. v(x) = T0 + r(2a − x)x/2.

9. Du′′ − Su′ = 0, 0 < x < a; u(0) = U , u(a) = 0,

u(x) = U(eSx/D − eSa/D)/(1 − eSa/D).


Section 2.31. w(x, t) = − 2

π(T0 + T1) sin

(πx

a

)exp

(−π2kt

a2

)

− 2

π

(T0 − T1

2

)sin

(2πx

a

)exp

(−4π2kt

a2

)− · · ·

3. The partial differential equation is

∂2U

∂ξ 2= ∂U

∂τ, 0 < ξ < 1, 0 < τ.

5. w(x, t) =∞∑

n=1

bn sin

(nπx

a

)exp

(−n2π2kt/a2), bn = T0

2(1 − cos(nπ))

πn.

7. w(x, t) as in the answer to Exercise 5, with bn = 2βa

π· 1

n.

9. a. v(x) = C1;

b.∂w

∂t= D

∂2w

∂x2, 0 < x < a, 0 < t,

w(0, t) = 0, w(a, t) = 0, 0 < t,

w(x,0) = C0 − C1;

c. C(x, t) = C1 +∞∑

n=1

bn sin

(nπx

a

)exp

(−n2π2kt/a2),

bn = (C0 − C1)2(1 − cos(nπ))

πn;

d. t = −a2

Dπ2ln

(π

40

);

e. t = 6444 s = 107.4 min.

Section 2.41. a0 = T1/2, an = 2T1(cos(nπ) − 1)/(nπ)2.

3. u(x, t) as given in Eq. (9), with λn = nπ/a, a0 = T0/2, and an =4T0(2 cos(nπ/2) − 1 − cos(nπ))/n2π2.

5. a. The general solution of the steady-state equation is v(x) = c1 + c2x.The boundary conditions are c2 = S0, c2 = S1; thus there is a solutionif S0 = S1. If heat flux is different at the ends, the temperature cannotapproach a steady state. If S0 = S1, then v(x) = c1 + S0x, c1 undefined.

Chapter 2 457

c. A = (S1 − S0)/a, B = S0. If S0 �= S1, then∂u

∂t= kA for all t.

7. φ′′ + λ2φ = 0, 0 < x < a,

φ(0) = 0, φ(a) = 0.

Solution: φn = sin(λnx), λn = nπ/a(n = 1,2, . . .).

9. The series∞∑

n=1

|An(t1)| converges.

11. No. u(0, t) is constant if ut(0, t) = 0.

Section 2.51. v(x, t) = T0.

3. u(x, t) = T0 +∞∑

n=1


nkt), λn = (2n − 1)π/2a,

bn = 8T(−1)n+1

π2(2n − 1)2− 4T0

π(2n − 1).

5. The steady-state solution is v(x) = T0 − Tx(x − 2a)/2a2. The transientsatisfies Eqs. (5)–(8) with

g(x) = T0 − v(x) = Tx(x − 2a)

2a2.

7. u(x, t) = T0 +∞∑

n=1

cn cos(λnx) exp(−λ2

nkt),

λn = (2n − 1)π/2a, cn = 4(T1 − T0)(−1)n+1

π(2n − 1).

9. u(x, t) = T1 cos(πx/2a) exp

(−

(π

2a

)2

kt

).

11. The graph of G in the interval 0 < x < 2a is made by reflecting the graphof g in the line x = a (like an even extension).

13. a. u(x, t) = T0 +∞∑

n=1


nkt), λn = (2n − 1)π/2a,

bn = 1

a

∫ 2a

0g(x) sin

(nπx

2a

)dx.


In the integral for bn, break the interval of integration at a; in the secondintegral, make the change of variable y = 2a−x. The two integrals cancelif n is even, and the coefficient is the same as Eq. (18) if n is odd.

b. In the solution of Eqs. (1)–(4), the eigenfunction φ(x) = sin((2n −1)πx/2a) has the property φ(2a − x) = φ(x), so the sum of the serieshas the same property. This implies 0 derivative at x = a.

15. W(t) = C0LA

[1 −

∞∑n=1

2e−(λ2nDt)

(n − 1/2)2π2

].

Section 2.61. The graph of v(x) is a straight line from T0 at x = 0 to T∗ at x = a, where

T∗ = T0 + ha

k + ha(T1 − T0).

In all cases, T∗ is between T0 and T1.

3. Negative solutions provide no new eigenfunctions.

7. bm = 2(1 − cos(λma))

λm[a + (κ/h) cos2(λma)].

9. bm = −2(κ + ah) cos(λma)

λm(ah + κ cos2(λma)).

Section 2.71. λn = nπ/ ln 2, φn = sin(λn ln(x)).

3. a. sin(λnx), λn = (2n − 1)π/2a;

b. cos(λnx), λn = (2n − 1)π/2a;

c. sin(λnx), λn a solution of tan(λa) = −λ;

d. λn cos(λnx) + sin(λnx), λn a solution of cot(λa) = λ;

e. λn cos(λnx) + sin(λnx), λn a solution of tan(λa) = 2λ/(λ2 − 1).

5. The weight functions in the orthogonality relations and limits of integra-tion are:

a. 1 + x, 0 to a; b. ex, 0 to a; c. 1x2 , 1 to 2; d. ex, 0 to a.

7. Because λ appears in a boundary condition.

9. The negative value of µ does not contradict Theorem 2 because the coef-ficient α2 is not positive.

Chapter 2 459

Section 2.8

1. x =∞∑

n=1

cnφn, 1 < x < b; cn = 2nπ1 − b cos(nπ)

n2φ2 + ln2(b)

.

3. 1 =∞∑

n=1

cnφn, 0 < x < a; cn = 2nπ1 − ea/2 cos(nπ)

n2π2 + a2/4.

(Hint: Find the sine series of ex/2.)

5. bn =∫ r

lf (x)ψn(x)p(x)dx.

7. 1 and√

2 cos(nπx), n = 1,2, . . . .

Section 2.91. a. v(x) = constant; b. v(x) = AI(x) + B.

3. If ∂u/∂x = 0 at both ends, then the steady-state problem is indeterminate.But Eqs. (1)–(3) are homogeneous, so separation of variables applies di-rectly. Note that λ0 = 0 and φ0 = 1. The constant term in the series foru(x, t) is

a0 =∫ r

l p(x)f (x)dx∫ rl p(x)dx

.

Section 2.101. The solution is as in Eq. (9), with B(λ) = 2T(cos(λa) − cos(λb))/λπ .

3. u(x, t) is given by Eq. (6) with B(λ) = 2T0λ

π(α2 + λ2).

5. u(x, t) =∫ ∞

0A(λ) cos(λx) exp

(−λ2kt)dλ;

A(λ) = 2T

πλ

(sin(λb) − sin(λa)

).

7. u(x, t) = T0 +∫ ∞

0B(λ) sin(λx) exp

(−λ2kt)dλ;

B(λ) = 2

π

∫ ∞

0

(f (x) − T0

)sin(λx)dx.

9. a. v(x) = C0e−ax;

b.∂w

∂t= D

(∂2w

∂x2− a2w

), 0 < x, 0 < t,


w(0, t) = 0, 0 < t,

w(x,0) = −C0e−ax, 0 < x;

c. w(x, t) = e−a2Dt

∫ ∞

0B(λ) sin(λx)e−λ2Dt dλ,

B(λ) = −2C0λ/(π

(λ2 + a2

)).

Section 2.111. Break the interval of integration at x′ = 0.

3. B(λ) = 0, A(λ) = 2T0a

π(1 + λ2a2).

5. The function u(x, t), as a function of x, is the famous “bell-shaped” curve.The smaller t is, the more sharply peaked the curve.

7. In Eq. (3) replace both f (x′) and u(x, t) by 1.

9. Using the integral given, obtain

u(x, t) = 2

π

∫ ∞

0

1

λsin(λx)e−λ2kt dλ.

Note, however, that B(λ) = 2/λπ is not found using the usual formulasfor Fourier coefficient functions.

Section 2.125. As t → 0+, x/

√4πkt →

{+∞ if x > 0,−∞ if x < 0,

so erf(x/√

4πkt ) →{+1 if x > 0,

−1 if x < 0.

7. Make the substitution x = y2. Then I(x) = √π erf(

√x) + c.

9. Let z be defined by erf(z) = −Ub/(Ui − Ub). Then x(t) = z√

4kt.

Chapter 2 Miscellaneous Exercises1. SS: v(x) = T0, 0 < x < a.

EVP: φ′′ + λ2φ = 0, φ(0) = 0, φ(a) = 0, λn = nπ/a, φn = sin(λnx),n = 1,2, . . . .

Chapter 2 461

u(x, t) = T0 +∞∑1

bn sin(λnx)e−λ2nkt ,

bn = 2

a

∫ a

0(T1 − T0) sin

(nπx

a

)dx.

3. SS: v(x) = T0 + r

2x(x − a), 0 < x < a.

EVP: φ′′ + λ2φ = 0, φ(0) = 0, φ(a) = 0, λn = nπ/a, φn = sin(λnx),n = 1,2, . . . .

u(x, t) = T0 − r

2x(x − a) +

∞∑1


nkt),

bn = 2

a

∫ a

0

[T1 − T0 + r

2x(x − a)

]sin

(nπx

a

)dx.

5. SS: not needed.

(Hint: Put −γ 2u on the other side of the equation. Separation of vari-ables gives φ′′/φ = γ 2 + T ′/kT = −λ2.)

EVP: φ′′ + λ2φ = 0, φ′(0) = 0, φ′(a) = 0, λ0 = 0, φ0 = 1; λn = nπ/a,φn = cos(λnx), n = 1,2, . . . .

u(x, t) = e−γ 2kt(

a0 +∑


nkt))

.

a0 = T1/2, an = −2T1

(1 − cos(nπ)

)/n2π2.

7. u(x, t) = T0.

9. u(x, t) = T0 +∞∑

n=1

cn sin(λnx) exp(−λ2

nkt),

λn = (2n − 1)π

2a, cn = (T1 − T0) · 4

(2n − 1)π.

11. u(x, t) = T0 +∫ ∞

0B(λ) sin(λx) exp

(−λ2kt)dλ, B(λ) = −2λT0

π(α2 + λ2).

13. u(x, t) =∫ ∞

0A(λ) cos(λx) exp

(−λ2kt)dλ, A(λ) = 2T0 sin(λa)

πλ.

15. u(x, t) =∫ ∞

0


)exp

(−λ2kt)dλ,

A(λ) = T0 sin(λa)

πλ, B(λ) = T0(1 − cos(λa))

πλor


u(x, t) = T0√4πkt

∫ a

0exp

(− (x′ − x)2

4kt

)dx′

= T0

2

[erf

(a − x√

4kt

)+ erf

(x√4kt

)].

17. Interpretation: u is the temperature in a rod with insulation on the cylin-drical surface and on the left end. At the right end, heat is being forcedinto the rod at a constant rate (because q(a, t) = −κ ∂u

∂x (a, t) = −κS, soheat is flowing to the left, into the rod). The accumulation of heat energyaccounts for the steady increase of temperature.

19. (1/6ka)u3 − (a/6k)u1 satisfies the boundary conditions.

21. w(x, t) = −2

u

∂u

∂x, where u(x, t) = a0 +

∑an cos(nπx) exp

(−n2π2t),

where a0 = 2(1 − e−1/2

)and an = 1 − e−1/2 cos(nπ)

14 + (nπ)2

.

23. u2 = T0β2V

β1 + β2, u1 = T0

(1 − β1V

β1 + β2

),

where V = 1 − exp(−(β1 + β2)t) and βi = h/ci.

25. u(ρ, t) = 1

ρ

∞∑n=1

bn sin(λnρ) exp(−λ2

nkt),

λn = nπ/a, bn = 2

a

∫ a

0ρT0 sin(λnρ)dρ.

27. v(x) = T0 + Sx − Ssinh(λx)

γ cosh(γ a).

29. If λ = 0, the differential equation is φ′′ = 0 with general solution φ(x) =c1 + c2x. The boundary conditions require c2 = 0 but allow c1 �= 0. Thus,this value of λ permits the existence of a nonzero solution, and thereforeλ = 0 is an eigenvalue.

31. Choose B(ω) = 2π

∫ ∞0 f (t) sin(ωt)dt. If f has a Fourier integral represen-

tation, then this choice of B will make u(0, t) = f (t), 0 < t.

33. a. v(x) = −Ix/aK + c1 + c2(1 − e−aKx/T),

c1 = h1, c2 = (h2 − h1 + IL/aK)/(1 − e−aKL/T).

b.∂2w

∂x2+ µ

∂w

∂x= 1

k

∂w

∂t, 0 < x < L, 0 < t,

w(0, t) = 0, w(L, t) = 0, 0 < t,

w(x,0) = h0(x) − v(x), 0 < x < L,

where µ = aK/T, k = T/S.

Chapter 3 463

c. w(x, t) =∑

cnφn(x)e−λ2nkT , φn(x) = e−µx/2 sin(nπx/L),

λ2n =

(nπ

L

)2

+ µ2

4;

d. λ2n = (7.30n2 + 0.0133) × 10−4 m−1.

35. a.∂u

∂t= D

∂2u

∂x2, 0 < x < L, 0 < t,

∂u

∂x(0, t) = 0, u(L, t) = S0, 0 < t;

u(0, t) = 0, 0 < x < L;

b. u(x, t) = S0 +∞∑

n=1

cn cos(λnx) exp(−λ2

nDt),

cn = 4S0(−1)n/(2n − 1).

37. T(y, t) = 300 − 150y/c + ∑bn sinλn(y + c) exp(−λ2

nkt), λn = nπ/2c,bn = (400 cos(nπ)+ 1000)/nπ . c. Just before time t = 0, the three termsadd to 0. Just after time t = 0, the integrated terms do not change sensi-bly, but in the first term, near y = c, T(y, t) changes suddenly.

Chapter 3

Section 3.11. [u] = L, [c] = L/t.

3. v(x) = (x2 − ax)g

2c2.

Section 3.2

3. u(x, t) =∞∑

n=1

bn sin

(nπx

a

)sin

(nπct

a

),

bn = 2a(1 − cos(nπ))

n2π2c.

5. u(x, t) =∞∑

n=1

an cos

(nπct

a

)sin

(nπx

a

), an = 2U0

1 − cos(nπ/2)

nπ.

7. a. sin

(nπx

a

); b. sin

(2n − 1

2

πx

a

).


9. Product solutions are φn(x)Tn(t), where

φn(x) = sin(λnx), Tn(t) = exp(−kc2t/2

) ×{

sin(µnt)cos(µnt),

λn = nπ

a, µn =

√λ2

nc2 − 1

4k2c4.

11. Product solutions are φn(x)Tn(t), where

φn(x) = sin

(nπx

a

),

Tn(t) = sin or cos

(n2π2ct

a2

).

Frequencies n2π2c/a2.

13. The general solution of the differential equation is φ(x) = A cos(λx) +B sin(λx) + C cosh(λx) + D sinh(λx). Boundary conditions at x = 0 re-quire A = −C, B = −D; those at x = a lead to C/D = −(cosh(λa) +cos(λa))/(sinh(λa)− sin(λa)) and 1 + cos(λa) cosh(λa) = 0. The firsteigenvalues are λ1 = 1.875/a, λ2 = 4.693/a, and the eigenfunctions aresimilar to the functions shown in the figure.

15. u(x, t) =∞∑

n=1

(an cos(µnt) + bn(sinµnt)

)sin(λnx): λn = nπ/a,

µn = √λ2

n + γ 2c, an = 2h(1 − cos(nπ)

)/nπ,bn = 0, n = 1,2, . . . .

17. Convergence is uniform because∑ |bn| converges.

Section 3.31. Table shows u(x, t)/h.

t

x 0 0.2a/c 0.4a/c 0.8a/c 1.4a/c0.25a 0.5 0.5 0.2 −0.5 −0.20.5a 1.0 0.6 0.2 −0.6 −0.2

3. u(0,0.5a/c) = 0; u(0.2a,0.6a/c) = 0.2αa; u(0.5a,1.2a/c) = −0.2αa.(Hint: G(x) = αx, 0 < x < a.)

5. G(x) ={0, 0 < x < 0.4a,

5(x − 0.4a), 0.4a < x < 0.6a,a, 0.6a < x < a.

Notice that G is a continuous function whose graph is composed of linesegments.

Chapter 3 465

Figure 2 Solution for Exercise 7, Section 3.3.


7. See Fig. 2.

9. See Fig. 3.

11. By the chain rule we calculate

∂u

∂x= ∂v

∂w

∂w

∂x+ ∂v

∂z

∂z

∂x= ∂v

∂w+ ∂v

∂z,

∂2u

∂x2= ∂

∂w

(∂v

∂w+ ∂v

∂z

)∂w

∂x+ ∂

∂z

(∂v

∂w+ ∂v

∂z

)∂z

∂x

= ∂2v

∂w2+ 2

∂2v

∂w∂z+ ∂2v

∂z2


and similarly

∂2u

∂t2= c2

(∂2v

∂w2− 2

∂2v

∂z∂w+ ∂2v

∂z2

).

(We have assumed that the two mixed partials ∂2v/∂z∂w and ∂2v/∂w∂zare equal.) If u(x, t) satisfies the wave equation, then

∂2u

∂x2= 1

c2

∂2u

∂t2.

In terms of the function v and the new independent variables this equa-tion becomes

∂2v

∂w2+ 2

∂2v

∂z∂w+ ∂2v

∂z2= ∂2v

∂w2− 2

∂2v

∂z∂w+ ∂2v

∂z2

or, simply,

∂2v

∂z∂w= 0.

13. u(x, t) = −c2 cos(t) + φ(x − ct) + ψ(x + ct).

Section 3.41. If f and g are sectionally smooth and f is continuous.

3. The frequency is cλn rads/sec, and the period is 2π/cλn sec.

5. Separation of variables leads to the following in place of Eqs. (11)and (12):

T ′′ + γ T ′ + λ2c2T = 0, (11′)(s(x)φ′)′ − q(x)φ + λ2p(x)φ = 0. (12′)

The solutions of Eq. (11′) all approach 0 as t → ∞, if γ > 0.

7. The period of Tn(t) = an cos(λnct) + bn sin(λnct) is 2π/λnc. All Tn’s havea common period p if and only if for each n there is an integer m suchthat m(2π/λnc) = p, or m = (pc/2π)λn is an integer. For λn as shownand β = q/r, where q and r are integers, this means

m =(

pc

2π

)α

(n + q

r

)

or

m =(

pc

2π

)α

r(rn + q).

Chapter 3 467


Given α, p can be adjusted so that m is an integer whenever n is an integer.

Section 3.51. If q ≥ 0, the numerator in Eq. (3) must also be greater than or equal to 0,

since φ1(x) cannot be identically 0.

3. 2π2/3 is one estimate from y = sin(πx).

5.

∫ 2

1(y′)2dx = 1

3,

∫ 2

1

y2

x4dx = 25

6− 6 ln 2;

N(y)/D(y) = 42.83; λ1 ≤ 6.54.

Section 3.61. u(x, t) = 1

2 [ fe(x + ct) + Go(x + ct)] + 12 [ fe(x − ct) − Go(x − ct)], where fe

is the even extension of f and Go is the odd extension of G.

3. See Fig. 4.

5. See Fig. 5.

7. u(x, t) = 1

2

[f (x + ct) + f (x − ct)

] + 1

2c

∫ x+ct

x−ctg(y)dy.


1. u(x, t) =∞∑1

bn sin(λnx) cos(λnct), bn = 2(1−cos(nπ)

)/nπ , λn = nπ/a.



Figure 6 Solution of Miscellaneous Exercise 3, Chapter 3.

3. See Fig. 6.

5. See Fig. 7.

7. See Fig. 8.

9. See Fig. 9.

11. See Fig. 10.

Chapter 3 469



13. See Fig. 11.

15. Using y(x) = x(1 − x), find λ21 ≤ 10.5.

17. f (q) = 12a2sech2(aq), c = 4a2.



Figure 10 Solution for Miscellaneous Exercise 11, Chapter 3.


21. v(x, t) =∞∑

n=1

(an cos(λnct) + bn sin(λnct)

)sin(λnx),

λn = (2n − 1)π/2a,

Chapter 4 471

an = 8aU0(−1)n+1

π2(2n − 1)2, bn = 0.

23.Y ′′

Y= 2V

k

ψ ′

ψ. The function φ(x − Vt) cancels from both sides.

25. φn(−Vt) = T0 exp(λ2nkt/2)bn, t > 0,

φn(x) = T1 exp(λ2nkx/2V)bn, x > 0,

where∞∑

n=1

bn sin(λny) = 1, 0 < y < b.

27. φ(x − ct) = e−c(x−ct)/k = e(c2t−cx)/k. The given c satisfies c2 = iωk,so φ(x − ct) = eiωt−(1+i)px = e−pxei(ωt−px). Now form 1

2 (φ(x − ct) +φ(x − ct)) = e−px cos(ωt − px) and so forth.

29. Differentiate and substitute.

31. φ(2) − εφ(4) + λ2φ = 0,

φ(0) = 0, φ(a) = 0,

φ′′(0) = 0, φ′′(a) = 0.

33. λn = nπ

a

√1 + ε

(nπ

a

)2 ∼= nπ

a.

Chapter 4

Section 4.11. f + d = 0.

3. Y(y) = A sinh(πy), A = 1/ sinh(π).

5. v(r) = a ln(r) + b.

7.∂u

∂x= ∂v

∂rcos(θ) − ∂v

∂θ

sin(θ)

r,

∂u

∂y= ∂v

∂rsin(θ) + ∂v

∂θ

cos(θ)

r.

9. a.∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, 0 < y < b,

u(0, y) = 0, u(a, y) = 0, 0 < y < b,

u(x,0) = f (x), u(x,b) = f (x), 0 < x < a.


Membrane is attached to a frame that is flat on the left and right but hasthe shape of the graph of f (x) at top and bottom.

b.∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, 0 < y < b,

∂u

∂x(0, y) = 0, u(a, y) = 0, 0 < y < b,

u(x,0) = 0, u(x,b) = 100, 0 < x < a.

The bar is insulated on the left; the temperature is fixed at 100 on the top,at 0 on the other two sides.

c.∂2u

∂x2+ ∂2u

∂y2= 0, 0 < x < a, 0 < y < b,

u(0, y) = 0, u(a, y) = 100, 0 < y < b,

∂u

∂y(x,0) = 0,

∂u

∂y(x,b) = 0, 0 < x < a.

The sheet is electrically insulated at top and bottom. The voltage is fixedat 0 on the left and 100 on the right.

d.∂2φ

∂x2+ ∂2φ

∂y2= 0, 0 < x < a, 0 < y < b,

∂φ

∂x(0, y) = 0,

∂φ

∂x(a, y) = −a, 0 < y < b,

∂φ

∂y(x,0) = 0,

∂φ

∂y(x,b) = b, 0 < x < a.

The velocities, given by V = −∇φ, are Vx = a, Vy = 0 on the right,Vx = 0, Vy = −b on the top; and walls on the other two sides make ve-locities 0 there.

Section 4.21. Show by differentiating and substituting that both are solutions of the

differential equation. The Wronskian of the two functions is∣∣∣∣ sinh(λy) sinh(λ(b − y))λ cosh(λy) −λ cosh(λ(b − y))

∣∣∣∣ = −λ sinh(λb) �= 0.

3. In the case b = a, use two terms of the series: u(a/2,a/2) = 0.32.

5. u(x, y) =∞∑1

bn sin

(nπx

a

)sinh(nπy/a)

sinh(nπb/a), bn = 8

n2π2sin

(nπ

2

).

Chapter 4 473

7. a. See Eq. (11). an = 0, cn = 200(1 − cos(nπ))/nπ ;

b. u(x, y) = u1(x, y) + u2(x, y),u1(x, y) is the solution to Part a,

u2(x, y) =∞∑

n=1

cnsinh(µnx)

sinh(µna)sin(µny),

µn = nπ/b, cn = 200(1 − cos(nπ))/nπ .

c. u(x, y) = u1(x, y) + u2(x, y), where

u1(x, y) =∞∑

n=1

cnsinh(λny)

sinh(λnb)sin(λnx),

u2(x, y) =∞∑

n=1

cnsinh(µnx)

sinh(µna)sin(µny).

In both series, cn = 2ab(−1)n+1/nπ . Also note u(x, y) = xy.

Section 4.31. a. u(x, y) = 1, but the form found by applying the methods of this sec-

tion is

u(x, y) =∞∑

n=1

ansinh(λny) + sinh(λn(b − y))

sinh(λnb)cos(λnx)

+∞∑

n=1

bncosh(µnx)

cosh(µna)sin(µny),

where

λn = (2n − 1)π

2a, an = 4 sin

((2n−1)π

2

)π(2n − 1)

,

µn = nπ

b, bn = 2(1 − cos(nπ))

nπ.

b. u(x, y) = y/b, and this is found by the methods of this section. In thiscase, 0 is an eigenvalue.

c.4

π

∞∑1

(−1)n+1 cos(λny)

(2n − 1)

sinh(λn(a − x))

sinh(λna), λn =

(2n − 1

2

π

b

).

3. b0b = V0

2, bn sinh(λnb) = 2V0(cos(nπ) − 1)

n2π2.


5. Check zero boundary conditions by substituting. At x = a, find

An cosh(µna) = 2

b

∫ b

0Sy cos(µny)dy.

7. w(x, y) =∞∑

n=1

an cosh(λny) cos(λnx). From the condition at y = b,

an cosh(λnb) = 2

a

∫ a

0

Sb

a(x − a) cos(λnx)dx.

9. w(x, y) =∞∑

n=1

cn sinh(λny) + an sinh(λn(b − y))

sinh(λnb)sin(λnx),

an = cn = −2

a

∫ a

0Hx(a − x) sin(λnx)dx = −2Ha2 1 − cos(nπ)

n3π3.

11. 12A + 2C = −K , 12E + 2C = −K . There are many solutions.

Section 4.41. an = 2

a

∫ a

0f (x) sin

(nπx

a

)dx.

3. A(µ) = 2

π

∫ ∞

0g2(y) sin(µy)dy.

5. a. u(x, y) =∑

cn cos(λnx) exp(−λny), λn = (2n − 1)π/2a,

cn = 4(−1)n+1/π(2n − 1);

b. u(x, y) =∫ ∞

0B(λ) cosh(λx) sin(λy)dλ, B(λ) = 2λ

π(λ2 + 1) cosh(λa);

c. u(x, y) =∫ ∞

0A(λ) cos(λy) sinh(λx)dλ, A(λ) = 2 sin(λb)

πλ sinh(λa).

7. u(x, y) =∞∑1

bn sin(λnx) exp(−λny)

+∫ ∞

0

(A(µ)

sinh(µx)

sinh(µa)+ B(µ)

sinh(µ(a − x))

sinh(µa)

)sin(µy)dµ,

λn = nπ/a, bn = 2(1 − cos(nπ))/nπ , A(µ) = B(µ) = 2µ/π(µ2 + 1).

Also see Exercise 8.

9. a. u(x, y) = 2

π

∫ ∞

0

1 − cos(λa)

λsin(λx)

sinh(λy)

sinh(λb)dλ;

Chapter 4 475

b. u(x, y) = 2

π

∫ ∞

0

λ

1 + λ2sin(λx)

sinh(λ(b − y))

sinh(λb)dλ.

11. u(x, y) =∫ ∞

0

2

π(1 + λ2)

sinh(λx)

sinh(λa)cos(λy)dλ.

13. e−λy sin(λx), λ > 0.

15. e−λy sin(λx), e−λy cos(λx), λ > 0.

17. u(x, y) = 1

π

[π

2+ tan−1(x/y)

].

19. This solution is unbounded as x tends to infinity and cannot be foundby the method of this section.

Section 4.51. v(r, θ) is given by Eq. (10) with bn = 0, a0 = π/2,

an = −2 (1 − cos(nπ))/πn2cn.

3. The solution is as in Eq. (10) with bn = 0, a0 = 1/π , a1 = 1/2, and

an = 2 sin((n − 1)π/2)

π(n2 − 1)for n �= 1.

5. Convergence is uniform in θ .

7. a0 = 1

2π

∫ π

−π

f (θ)dθ , an = cn

π

∫ π

−π

f (θ) cos(nθ)dθ ,

bn = cn

π

∫ π

−π

f (θ) sin(nθ)dθ .

9.2

π

∞∑n=1

1 − cos(nπ)

nc2nr2n sin(2nθ) = v(r, θ).

11. vn(r, θ) = rn/α sin(nθ/α) has ∂v/∂r unbounded as r → 0+, if n = 1.

Section 4.61. Hyperbolic (a) and (e); elliptic (b) and (c); parabolic (d).

3. Only (e).

5. a. u(x, y) =∞∑1

an sin(nπx)e−nπy;

b. u(x, y) =∞∑1

an sin(nπx) cos(nπy);


c. u(x, y) =∞∑1

an sin(nπx) exp(−n2π2y),

an = 2

∫ 1

0f (x) sin(nπx)dx.

7. X′′/X = −λ2, T ′′/T = −λ2/(1 + ελ2).


1. u(x, y) =∞∑1

bnsinh(λn(a − x))

sinh(λna)sin(λny),

λn = nπ/b, bn = 2(1 − cos(nπ))/nπ .

3. u(x, y) = 1. Note that 0 is an eigenvalue.

5. u(x, y) =∞∑

n=1

an sinh(λnx) + bn sinh(λn(a − x))

sinh(λna)cos(λny),

λn = (2n − 1)π/2b, an = bn = 4(−1)n+1/π(2n − 1).

7. u(x, y) = w(x, y) + w(y, x), where

w(x, y) =∞∑

n=1

bnsinh(λn(a − y))

sinh(λna)sin(λnx),

λn = nπ/a, bn = 8h

n2π2sin

(nπ

2

).

9. u(x, y) =∫ ∞

0A(λ)

sinh(λ(b − y))

sinh(λb)cos(λx)dλ, A(λ) = 2 sin(λa)/λπ .

11. u(x, y) =∫ ∞

0A(λ) cos(λx)e−λydλ, A(λ) = 2α/π

(α2 + λ2

).

13. u(x, y) = −1

πtan−1

(x − x′

y

)∣∣∣∣∞

−∞= 1

π

[π

2−

(−π

2

)].

15. u(r, θ) = a0 +∞∑

n=1

(r

c

)n(an cos(nθ) + bn sin(nθ)

),

a0 = 1

2, an = 0, bn = 1 − cos(nπ)

nπ.

17. Same form as Exercise 15, but a0 = 2/π ,

an = 2(1 + cos(nπ))/(1 − n2), bn = 0 (and a1 = 0).

19. u(r, θ) = (ln(r) − ln(b))/(ln(a) − ln(b)).

Chapter 4 477

21. u(r, θ) =∞∑1

bn

(r

c

)n/2

sin(nθ/2), bn = 1

π

∫ 2π

0f (θ) sin(nθ/2)dθ .

23. u(x, y) =∑

cn sinh(λny) sin(λnx), λn = (2n − 1)π/2a,

cn = 2 sin(λna)/(aλ2n sinh(λnb)).

25. w satisfies the potential equation in the rectangle with boundary condi-tions

w(0, y) = 0, wx(a, y) = ay/b, 0 < y < b,

w(x,0) = 0, w(x,b) = 0, 0 < x < a.

w(x, y) =∞∑

n=1

bn sin(λny) cosh(λnx),

λn = nπ/b, bn = 2a(−1)n+1/n2π2 cosh(λna).

27. The equations become

∂2φ

∂y∂x= ∂2φ

∂x∂y,

(1 − M2

)∂2φ

∂x2+ ∂2φ

∂y2= 0.

29. φ(x, y) =∫ ∞

0

(A(α) cos(αx) + B(α) sin(αx)

)e−βydα + c,

where β = α√

1 − M2, c is an arbitrary constant, and

A(α)

B(α)

}= − U0

βπ

∫ ∞

−∞f ′(x)

{cos(αx)sin(αx)

}dx.

31. If (x(s), y(s)) is the parametric representation for the boundary curve C,then the vector y′i − x′j is normal to C, and∫

C

∂u

∂nds =

∫C

∂u

∂xdy − ∂u

∂ydx.

By Green’s theorem,

∫C

∂u

∂xdy − ∂u

∂ydx =

∫∫R

(∂2u

∂x2+ ∂2u

∂y2

)dA,

which is 0, since u satisfies the potential equation in R.

33. Substitute directly.

35. −∇u = −(xi + yj)/(x2 + y2).


37. a. u = − r2

4+ c1 ln(r) + c2;

b. u = − (ln(r))2

2+ c1 ln(r) + c2.

39. V(x, y) ∼= a0 = 1

a

∫ a

0f (x)dx if y > 5L (because e−λ1.5L ∼= 0).

41. The solution is θ(X,Y) = 1. The low Biot number, B = 0, means a verylarge value for conductivity κ , so very little or no cooling takes place.

Chapter 5

Section 5.11.

∂2u

∂x2+ ∂2u

∂y2= 1

c2

∂2u

∂t2, 0 < x < a, 0 < y < b, 0 < t,

u(x,0, t) = 0, u(x,b, t) = 0, 0 < x < a, 0 < t,

u(0, y, t) = 0, u(a, y, t) = 0, 0 < y < b, 0 < t,

u(x, y,0) = f (x, y),∂u

∂t(x, y,0) = g(x, y), 0 < x < a, 0 < y < b.

3.∂2u

∂x2+ ∂2u

∂y2+ ∂2u

∂z2= 1

c2

∂2u

∂t2.

Section 5.21.

∫ c

0

∂2u

∂z2dz = ∂u

∂z

∣∣∣∣c

0

= 0 by Eq. (12).

3. W ′′ + (2h/bκ)(T2 − W) = 0, 0 < x < a, W(0) = T0, W(a) = T1.

W(x) = T2 + A cosh(µx) + B sinh(µx), where µ2 = 2h/bκ ,

A = T0 − T2, B = (T1 − T2 − A cosh(µa))/ sinh(µa).

5. ∇2u = 1

k

∂u

∂t, 0 < x < a, 0 < y < b, 0 < t,

∂u

∂x(0, y, t) = 0, u(a, y, t) = T0, 0 < y < b, 0 < t,

u(x,0, t) = T0,∂u

∂y(x,b, t) = 0, 0 < x < a, 0 < t,

u(x, y,0) = f (x, y), 0 < x < a, 0 < y < b.

Chapter 5 479

Section 5.31. If a = b, the lowest eigenvalues are those with indices (m,n), in this or-

der: (1,1); (1,2) = (2,1); (2,2); (3,1) = (1,3); (3,2) = (2,3); (1,4) =(4,1); (3,3).

3. Frequencies are λmnc/2π (Hz), where λ2mn are the eigenvalues found in

the text.

5. λ2mn = (mπ/a)2 + (nπ/b)2, for m = 0,1,2, . . . , n = 1,2,3, . . . .

7. a. u(x, y, t) = 1.

For b and c the solution has the form

u(x, y, t) =∑m,n

amn cos

(mπx

a

)cos

(nπy

b

)exp

(−λ2mnkt

),

where λ2mn = (mπ/a)2 + (nπ/b)2 and m and n run from 0 to ∞.

b. a00 = (a + b)

2, am0 = −2b(1 − cos(mπ))

m2π2,

a0n = −2a(1 − cos(nπ))

n2π2, amn = 0 otherwise;

c. a00 = ab

4, am0 = −ab(1 − cos(mπ))

m2π2, a0n = −ab(1 − cos(nπ))

n2π2,

amn = 4ab(1 − cos(nπ))(1 − cos(mπ))

m2n2π4

if m and n are greater than zero.

9. The choice of a positive constant for either X′′/X or Y ′′/Y , under theboundary conditions in Eqs. (9) and (10), will lead to the trivial solution.

11. The nodal lines form a grid: umn(x, y, t) = 0 at x = 0, a/m, 2a/m, . . . ,aand at y = 0, b/n, 2b/n, . . . ,b.

Section 5.41. The partial differential equations are the same, the boundary conditions

become homogeneous, and in the initial conditions g(r, θ) is replaced byg(r, θ) − v(r, θ).

3. In the heat problem, T ′ +λ2kT = 0. In the wave problem, T ′′ +λ2c2T = 0.

5. The boundary conditions Eqs. (10) and (11) would be replaced by

Q(0) = 0, Q(π) = 0.

Solutions are Q(θ) = sin(nθ), n = 1,2, . . . .


7. Taking the hint and using the fact that ∇2φ = −λ2φ, the left-hand sidebecomes (

λ2k − λ2

m

)∫∫R

φkφm

while the right-hand side is zero, because of the boundary condition.

Section 5.51. λn = αn/a, where αn is the nth zero of the Bessel function J0. The solutions

are φn(r) = J0(λnr) or any constant multiple thereof.

3. This is just the chain rule.

5. Rolle’s theorem says that if a differentiable function is zero in two places,its derivative is zero somewhere between. From Exercise 4 it is clear thatJ1 must be zero between consecutive zeros of J0. Check Fig. 7 and Table 1.

7. Use the second formula of Exercise 6, after replacing µ by µ + 1 on bothsides.

9. u(r) = T + (T1 − T)I0(γ r)/I0(γ a).

Section 5.61. v(0, t)/T0

∼= 1.602 exp(−5.78τ ) − 1.065 exp(−30.5τ ), where τ = kt/a2.

3. v(r, t) =∞∑

n=1

anJ0(λnr) exp(−λ2nkt), λn = αn/a. Use Eq. (13) and others to

find an = T0J1(αn/2)/αnJ21(αn).

5. Integration leads to the equality∫ a

0

(rφ′2)′

dr + λ2

∫ a

0r2

(φ2

)′dr = 0.

The first integral is evaluated directly. The second must be integrated byparts.

Section 5.71. Use

1

r

d

dr

(r

d

drJ0(λr)

)= −λ2J0(λr).

3. The frequencies of vibration are λmnc = αmnc/a. The five lowest values ofαmn, in order, have subscripts (0,1), (1,1), (2,1), (0,2), and (3,1). SeeTable 1 in Section 5.6.

Chapter 5 481

5. φ(a, θ) = 0 and φ(r, θ) = φ(r, θ + 2π).

7. Set Jm(λmnr) = φn. Then (rφ′n)

′ = −λmnrφn and (rφ′q)

′ = −λmqrφq are theequations satisfied by the functions in the integrand. Follow the proof inSection 2.7.

9. Generally, radii for φ0n are r = λ0m/λ0n for m = 1,2, . . . ,n. For n = 2,r = 2.405/5.520 = 0.436 and 1.

Section 5.81. φ(x) = xα[AJp(λx) + BYp(λx)], where α = (1 − n)/2, p = |α|.3. For λ2 = 0, Z = A + Bz.

5. φ(ρ + ct) = Fo(ρ + ct) + Ge(ρ + ct),

ψ(ρ − ct) = Fo(ρ − ct) − Ge(ρ − ct),

where Fo(x) is the odd periodic extension with period 2a of xf (x)/2 andGe(x) is the even periodic extension with period 2a of

∫(x/2c)g(x)dx.

7. The weight function is ρ2 and the interval is 0 to a.

9. v(x) = (b − x)(x − a)/(a + b)x2.

11. No. The idea is to find a solution of the partial differential equation thatdepends on only one variable. That is impossible if f depends on both xand y.

13. an = −∫ b

a v(x)Xn(x)x3dx∫ ba X2

n(x)x3dx.

Section 5.91. [k(k + 1) − µ2]ak+1 − [k(k − 1) − µ2]ak−1 = 0, valid for k = 1,2, . . . .

3. P5 = 1

8

(63x5 − 70x3 + 15x

).

5. y = A ln

(1 + x

1 − x

).

7. Differentiate Eq. (9) and add to it n times Eq. (8).

9. Leibniz’s rule states that

(uv)(k) =k∑

r=0

(kr

)u(k−r)v(r).


(A superscript k in parentheses means kth derivative.) The right-handside looks like the binomial theorem. In the case at hand, at most twoterms are not zero.

11. bn =

0 (n odd),−(−1)n/2(2n + 1)

(n + 2)(n − 1)

1 · 3 · 5 · · · (n − 1)

2 · 4 · 6 · · ·n(n even).

Section 5.101. The solution is as given in Eq. (5) with coefficients as shown in Eq. (7).

The integration yields (see Section 5.9)

b0 = 1

2; bn = 0 for n even; b1 = 3/4 and

bn = (−1)(n−1)/2

2 · cn

1 · 3 · 5 · · · (n − 2)

2 · 4 · 6 · · · (n − 1)· 2n + 1

n + 1, n = 3,5,7, . . . .

3. u(φ, t) = T − ∑bnPn(cos(φ)) exp(−(µ2 + n(n + 1))kt/R2), where µ2 =

γ 2R2, n is odd, and bn is as at the end of Part B, with T0 = T.

5. The eigenfunctions are as in Part C, except that n must be odd in order tosatisfy the boundary condition.

7. The nodal surfaces are: a sphere at ρ = 0.634 and two naps of a cone givenby φ = 0.304π and ρ = 0.696π .


1. u(x, y, t) =∞∑

m=1

am sin(µmy) exp(−µ2

mkt)

+∞∑

n=1

amn cos(λnx) sin(µmy) exp(−(

µ2m + λ2

n

)kt

),

µm = mπb, λn = nπ/a,

am = T1 − cos(mπ)

mπ, amn = 4T

π3

(cos(nπ) − 1)(1 − cos(mπ))

n2m.

3. u(a/2,b/2, t) =∞∑

n=1

bmn sin

(nπ

2

)sin

(mπ

2

)exp

(−(λ2

n + µ2m

)kt

),

where λn = nπ/a, µm = mπ/b, and

bmn = 4T

π2

(1 − cos(mπ))(1 − cos(nπ))

mn.

Chapter 5 483

The first three nonzero terms are, for a = b, those with (m,n) =(1,1), (1,3) = (3,1), (3,3). All terms with an even index are 0.

u(a/2,a/2, t) ∼= 16T

π2

(e−2τ − 2

3e−10τ + 1

9e−18τ

),

where τ = ktπ2/a2.

5. u(r) = (a2 − r2

)/2 and u(r) =

∞∑1

CnJ0(λnr), with Cn = 2a2

α3nJ1(αn)

.

7. w(x, t) = a0 +∞∑

n=1


nkt),

v(y, t) =∑

bm sin(µmy) exp(−µ2

mkt),

where µm = mπ/b, λn = nπ/a, and initial conditions are

v(y,0) = 1, 0 < y < b; w(x,0) = Tx/a, 0 < x < a.

9. J0(λr) exp(−λ2kt).

11. Bk = bk/k(k + 1) for k = 1,2, . . . ; b0 must be 0, and B0 is arbitrary.

13.((

1 − x2)y′)′ − m2

1 − x2y + µ2y = 0.

15. u(r, z) =∞∑

n=1

ansinh(λnz)

sinh(λnb)J0(λnr),

where λn = αn

aand an = 2U0

αnJ1(αn).

17. u(r, z, t) = sin(µz)J0(λr) sin(νct) is a product solution if µ = mπ/b, λ =αn/a, and ν = √

µ2 + λ2. The frequencies of vibration are therefore νcor

c

√(mπ

b

)2

+(

αn

a

)2

.

19. Each of the two terms satisfies ∇2φ = −(5π2)φ. On y = 0 and x = 1,both terms are 0; on y = x they are obviously equal in value, opposite insign.

21. Each term satisfies ∇2φ = −(16π2/3)φ.

On y = 0, φ = sin(2nπx) − sin(2nπx);

on y = √3x, φ = sin(4nπx) + 0 − sin(2nπ · 2x);


on y = √3(1 − x), φ = sin(4nπx) + sin(2nπ(1 − 2x)) − sin 2nπ .

23. For a sextant, φn = J3n(λr) sin(3nθ) and J3n(λ) = 0. Thus λ1 = 6.380,which is less than

√16π2/3 = 7.255.

25. b = −1.

27. Since y(x) = hJ0(kx)/J0(ka), where k = ω/√

gU , the solution cannothave this form if J0(ka) = 0.

29. u(r, t) = R(t)T(t); R(r) = r−mJm(λr), where m = (n − 2)/2; T(t) =a cos(λct) + b sin(λct).

31. b = DrL2

DzR2, ρ = UL

Dz.

33. a0 = 12.77, a1 = −4.88.

35. tan(λ) = Dλ/(D + λ2), λ = π (D = 0), 3.173 (D = 1), 4.132 (D = 10).

Chapter 6

Section 6.11. c.

s2 + 2ω2

s(s2 + 4ω2); d.

ω cos(φ) − s sin(φ)

s2 + ω2;

e.e2

s − 2; f.

2ω2

s(s2 + 4ω2).

3. a.e−as

s; b.

e−as − e−bs

s; c.

1 − e−as

s2.

5. a. e−t sinh(t); b. e−t cos(t); c. e−at sin(√

b2 − a2t)√b2 − a2

.

7. a.eat − ebt

a − b; b.

t

2asinh(at); d.

t2eat

2;

e. f (t) ={1, 0 < t < 1,

0, 1 < t.

9. a. [sin(ωt) − ωt cos(ωt)]/2ω2;

b. See Table 2;

c. See 7c;

d. [cos(ωt) − ωt sin(ωt)]/2.

Chapter 6 485

Section 6.21. a. e2t ; b. e−2t ; c.

3e−t − e−3t

2; d.

sin(3t)

3.

3. a.1 − e−at

a; b. t − sin(t); c.

sin(t) − 12 sin(2t)

3;

d. (sin(2t) − 2t cos(2t))/16; e. −3

4+ 1

2t + e−t − 1

4e−2t ; f. cosh(t) − 1.

5. a.e2t − e−2t

4; b.

1

2sin(2t);

c.3

2+ i

√2 − 3

4exp

(−i√

2t) − i

√2 − 3

4exp

(i√

2t); d. 4

(1 − e−t

).

7. a. 1 − cos(t); b.et − cos(ωt) + ω sin(ωt)

ω2 + 1; c. t − sin(t).

Section 6.3

1. a. s = −(

2n − 1

2π

)2

, n = 1,2, . . . ;

b. s = ±i2n − 1

2π , n = 1,2, . . . ;

c. s = ±inπ , n = 0,1,2, . . . ;

d. s = iη, where tanη = −1

η;

e. s = iη, where tanη = 1

η.

3. a.sinh(

√sx)

s2 sinh(√

s); b.

1

s− cosh

(√s( 1

2 − x))

s(s + 1) cosh(√

s/2).

5. a. u(x, t) = x +∞∑

n=1

2 sin(nπx)

nπ cos(nπ)exp

(−n2π2t);

b. u(x, t) is 1 minus the solution of Example 3.

Section 6.41. t + x2

2.

3. v(x, t) = 4

π2

∞∑1

cos((2n − 1)( 1

2 − x))

sin((2n − 1)π t)

(2n − 1)2 sin(

2n−12 π

) .


5. a.ω

ω2 − π2

(1

πsin(π t) − 1

ωsin(ωt)

)sin(πx);

b.1

2π2

(sin(π t) − π t cos(π t)

)sin(πx).

7. a. u(x, t) = x − sin(√

ax)

sin(√

a)e−at + 2a

π

∞∑1

sin(nπx) exp(−n2π2t)

n(a − n2π2) cos(nπ);

b. The term −x cos(nπx)

cos(nπ)exp

(−n2π2t)

arises.


1. U(s) = T0

γ 2 + s+ γ 2T

s(γ 2 + s),

u(x, t) = T0 exp(−γ 2t

) + T(1 − exp

(−γ 2t))

.

3. U(s) = cosh(√

sx)

s2 cosh(√

s),

u(x, t) = t − 1 − x2

2+

∞∑n=1

2 cos(ρnx)

ρ3n sin(ρn)

exp(−ρ2

nt),

where ρn = (2n−1)π

2 .

5. u(x, t) = x(1 − x)

2−

∞∑n=1

4 cos(ρn(x − 1

2 ))

ρn sin(ρn/2)exp

(−ρ2nt

),

where ρn = (2n − 1)π .

7. u(x, t) = x +∞∑1

2 sin(nπx)

nπ cos(nπ)exp

(−n2π2t).

9. U(x, s) = 1

s

(1 − exp

(−√sx

)).

11. f (t) = x√4π t3

exp

(−x2

4t

).

13. u(x, t) =∞∑

n=0

[erfc

(2n + 1 − x√

4t

)− erfc

(2n + 1 + x√

4t

)].

15. F(s) = 2∞∑

n=1

1

s2 + n2.

Chapter 7 487

17. f (t) =∞∑

−∞

1

2aG

(inπ

a

)einπ t/a.

19. F(s) must be of the form F(s) = G(s)/H(s), where G(s) is never infinite.The solutions of H(s) = 0 must form an arithmetic sequence of purelyimaginary numbers, and H′(s) �= 0 if H(s) = 0.

21. F(s) = (1 − e−π s)2

s(1 − e−2π s)= 1 − e−π s

s(1 + e−π s).

23. F(s) = 1 + e−π s

(s2 + 1)(1 − e−π s).

25. u(x, t) = sin(ωx) sin(ωt)

sin(ω)+

∞∑n=1

(−1)n+12ω

ω2 − n2π2sin(nπx) sin(nπ t).

27. U(x, s) = ω

s2 + ω2emx, m = 1

2−

√1

4+ s.

29. α2 = 1

2

(1

4±

√(1

4

)2

+ ω2

). Since α must be real, take the + sign.

Chapter 7

Section 7.11. 16(ui+1 − 2uij + ui−1) = −1, i = 1,2,3, u0 = 0, u4 = 1. Solution: u1 =

11/32, u2 = 5/8, u3 = 27/32.

3. 16(ui+1 − 2ui + ui−1) − ui = − 12 i, i = 1,2,3, u0 = 0, u4 = 1. Solution:

u1 = 0.285, u2 = 0.556, u3 = 0.800.

5. 16(ui+1 − 2ui + ui−1) = 14 i, i = 0,1,2,3, u0 − 2(u1 − u−1) = 1, u4 = 0.

Solution: u0 = 0.422, u1 = 0.277, u2 = 0.148, u3 = 0.051.

7. n = 3: u1 = 4.76, u2 = 4.24; n = 4: u1 = 6.65, u2 = 9.14, u3 = 5.92. Theactual solution,

u(x) = − sin(√

10x)

sin(√

10),

has a maximum of about 50. The boundary value problem is nearlysingular.


9. 25(ui+1 − 2ui + ui−1) − 25ui = −25, i = 1,2,3,4,5; u0 = 2, u5 + (u6 −u4)/(2/5) = 1. When the equation for i = 5 and the boundary conditionare combined, they become 2u4 − 3.4u5 = −1.4. Solution: u1 = 1.382,u2 = 1.146, u3 = 1.057, u4 = 1.023, u5 = 1.014.

11. 9(ui+1 − 2ui + ui−1) + (3/2)(ui+1 − ui−1) − ui = −(1/3)i, i = 0,1,2;u3 = 1, (u1 − u−1)/(2/3) = 0. When u−1 is eliminated and coefficientsare collected, the equations to solve are

−19u0 + 18u1 = 0,

7 12 u0 − 19u1 + 10 1

2 u2 = − 13 ,

7 12 u1 − 19u2 = −11 1

6 .

Solution: u0 = 0.795,u1 = 0.839,u2 = 0.919.

Section 7.21. Line m of the solution should be exactly the same as line m + 1 of Table 4.

3. r = 2/5, t = 1/40.

i

m 0 1 2 3 40 0 0 0 0 01 1 0. 0. 0. 0.

2 1 0.4 0. 0. 0.

3 1 0.48 0.16 0. 0.

4 1 0.56 0.224 0.064 0.

5 1 0.6016 0.2944 0.1024 0.0512

5. t = 1/32. Remember that u4(m) = u0(m) = mt. All numbers in thetable should be multiplied by t.

i

m 0 1 2 3 40 0 0 0 0 01 1 0 0 0 12 2 1/2 0 1/2 23 3 1 1/2 1 34 4 7/4 1 7/4 45 5 5/2 7/4 5/2 5

7. t = 1/32. All numbers in this table should be multiplied by t.

Chapter 7 489

i

m 0 1 2 3 40 0 0 0 0 01 0 1 1 1 02 0 3/2 2 3/2 03 0 2 5/2 2 04 0 9/4 3 9/4 0...∞ 0 3 4 3 0

9. t = 132 . Remember u−1 = u1. All entries in this table should be multi-

plied by x = 14 .

i

m 0 1 2 3 40 0 1 2 3 41 1 1 2 3 42 1 3/2 2 3 43 3/2 3/2 9/4 3 44 3/2 15/8 9/4 25/8 45 15/8 15/8 5/2 25/8 4

Section 7.3

1.i

m 0 1 2 3 40 0 0 0 0 01 0 1/4 1/4 1/4 02 0 1/4 1/2 1/4 03 0 1/4 1/4 1/4 04 0 0 0 0 05 0 −1/4 −1/4 −1/4 0

3. In this table, α = 1/√

2.

i

m 0 1 2 3 40 0 0 0 0 01 0 α/4 1/4 α/4 02 0 1/4 α/2 1/4 03 0 α/4 1/4 α/4 04 0 0 0 0 05 0 −α/4 −1/4 −α/4 0


5.i

tm m 0 1 2 3 40 0 0 1/2 1 1/2 00.177 1 0 1/2 3/4 1/2 00.354 2 0 3/8 1/4 3/8 00.530 3 0 0 −1/8 0 00.707 4 0 −7/16 −3/8 −7/16 00.884 5 0 −5/8 −11/16 −5/8 0

7.i

m 0 1 2 3 40 0 0 0 0 01 0 0 0 0 12 0 0 0 1 13 0 0 1 1 14 0 1 1 1 05 0 1 1 0 −16 0 0 0 −1 −17 0 −1 −2 −1 −18 0 −2 −2 −2 0

9. Run: ui(m + 1) = (2 − 2ρ2 − 16t2) ui(m) + ρ2ui−1(m) + ρ2ui+1(m) −ui(m − 1). Start: ui(1) = 1

2 ((2 − 2ρ2 − 16t2) ui(0) + ρ2ui−1(0) +ρ2ui+1(0)). Longest stable time step: t = 1/

√24(ρ2 = 2/3).

i

m 0 1 2 3 40 0 0.50 1.00 0.50 01 0 0.33 0.33 0.33 02 0 −0.28 −0.56 −0.28 03 0 −0.70 −0.70 −0.70 04 0 −0.19 −0.38 −0.19 05 0 0.45 0.45 0.45 06 0 0.49 0.98 0.49 07 0 0.21 0.21 0.21 08 0 −0.35 −0.71 −0.35 0

Section 7.41. At (1/4,1/4), 11/256; at (1/2,1/4), 14/256; at (1/2,1/2), 18/256.

3. In both this exercise and Exercise 4, the exact solution is u(x, y) = xy, andthe numerical solutions are exact.

Chapter 7 491

5. Coordinates and values of the corresponding ui are: (1/7,1/7), 5α;(2/7,1/7), 10α; (3/7,1/7), 14α; (1/7,2/7), 21α; (2/7,2/7), 32α. Hereα = 19/1159.

7. u1 = 0.670, u2 = 0.721, u3 = 0.961, u4 = 1.212, u5 = 0.954, u6 = 0.651.The remaining values are found by symmetry.

9. u1 = 0.386, u2 = 0.542, u3 = 0.784, u4 = 0.595. The remaining values arefound by symmetry.

Section 7.51. Use Eq. (8) with r = 1/4.

i

m 1 2 3 4 5 60 0 0 0 0 0 01 0 0 0 1/4 1/4 1/42 1/16 1/16 1/16 5/16 3/8 5/163 3/32 1/8 3/32 23/64 27/64 23/64

3. Note that u1 = u2 = u4 = u5; replacement equations become

u1(m + 1) = u3(m)/4, u3(m + 1) = u1(m).

i

m 1 30 1 11 1/4 12 1/4 1/43 1/16 1/44 1/16 1/16

5. Use Eq. (8) with r = 1/4. Note that u4 = u2, u7 = u3, u8 = u6.

i

m 1 2 3 5 6 90 0 0 0 0 0 01 0 0 1/4 0 1/4 1/22 0 1/16 5/16 1/8 7/16 5/83 1/32 7/64 3/8 1/4 17/64 23/32

7. Use the same numbering as for Exercise 5. Note that u1 = u3 = u7 = u9

and u2 = u4 = u6 = u8. The running equations become

u1(m + 1) = 1

2u2(m) − u1(m − 1),


u2(m + 1) = 1

2u1(m) + 1

4u5(m) − u2(m − 1),

u5(m + 1) = u2(m) − u5(m − 1).

m

0 1 2 3 4 5 6 7u1 0 0 0 1/8 0 −5/16 0 15/32u2 0 0 1/4 0 −3/8 0 5/16 0u5 0 1 0 −3/4 0 3/8 0 −1/16

9.i

m 1 2 50 1 1 11 1/2 3/4 12 −1/4 0 1/23 −1/2 −3/4 −3/24 −1/2 −5/4 −25 −3/4 −3/4 −1

11. See Fig. 12 below for numbering of points.

i

m 1 2 3 4 5 60 1 0 0 0 0 01 0 1/4 1/4 0 0 02 −3/4 0 0 1/4 1/8 03 0 −1/2 −7/16 0 0 3/16

Figure 12 Solution of Exercise 11, Section 7.5.

Chapter 7 493


1.ui+1 − 2ui + ui−1

(x)2− √

24xiui = 0, i = 0,1,2,

u1 − u−1

2x= 1, u3 = 1;

−18u0 + 18u1 = 6,

9u0 − 20.83u1 + 9u2 = 0,

9u1 − 22u2 = −9,

u0 = −0.248, u1 = 0.08, u2 = 0.44.

3.ui+1 − 2ui + ui−1

(x)2+ 1

1 + xi

ui+1 − ui−1

2x= −(1 + xi),

i = 1,2,3;u0 = 1,u4 = 0;

−32u1 + 17.60u2 = −15.65,

14.67u1 − 32u2 + 17.33u3 = −1.5,

14.86u2 − 32u3 = −1.75,

u1 = 0.822, u2 = 0.606, u3 = 0.335.

5. ui(m + 1) = (ui−1(m) + ui+1(m))/2. Note that u3(m) = u1(m) andu4(m) = u0(m).

i

m 0 1 20 0 0 01 0.03 0 02 0.06 0.015 03 0.09 0.03 0.154 0.12 0.053 0.035 0.14 0.075 0.0536 0.17 0.1 0.0757 0.20 0.122 0.108 0.22 0.15 0.122

7. First problem: ui(m+1) = (ui+1(m)+ui(m)+ui−1(m))/3; second prob-lem: ui(m + 1) = (ui+1(m) + ui−1(m))/3.


First Problem Second Problemi i

m 0 1 2 3 4 0 1 2 3 40 0 1 1 1 0 0 1 1 1 01 0 2/3 1 2/3 0 0 1/3 2/3 1/3 02 0 5/9 7/9 5/9 0 0 2/9 2/9 2/9 03 0 14/27 17/27 14/27 0 0 2/27 4/27 2/27 04 0 31/81 45/81 31/81 0 0 4/81 4/81 4/81 0

9. ui(m + 1) = (ui+1(m) + ui−1(m))/2.

i

m 0 1 2 3 4 50 0 0 0 0 0 01 1/2 0 0 0 0 02 1 1/4 0 0 0 03 3/2 1/2 1/8 0 0 04 2 13/16 1/4 1/16 0 05 5/2 9/8 7/16 1/8 1/32 06 3 87/32 5/8 15/64 1/16 0

11.i

m 0 1 2 3 40 0 0 0 0 01 0 0 0 0 12 0 0 0 1 13 0 0 1 1 14 0 1 1 1 15 0 1 1 1 16 0 0 1 1 17 0 0 0 1 18 0 0 0 0 1

13. Let uij∼= u(xi, yj). Then u11 = u22 = u33 = 0.5, u12 = 0.698, u13 = 0.792,

u21 = 0.302, u23 = 0.624, u21 = 0.209, u32 = 0.376.

15. Number as in Chapter 7, Fig. 4. Then u1 = u3 = u7 = u9 and u2 = u4 =u6 = u8.

m

0 1 2 3 4 5u1 1 1/2 3/8 1/4 3/16 1/8u2 1 3/4 1/2 3/8 1/4 3/16u5 1 1 3/4 1/2 3/8 1/4

Index

A

acoustic vibrations, 235

aluminum nitride, oxygen removalfrom, 156

antenna vibrations, 226

approximation by Fourier series, 91–94

arbitrary periods, 64–65

B

bad discontinuities, 74–75, 79

band-limited functions, 121–123

bars, insulated, 157–161. See also heatconduction problems

beam vibrations, 225

Bessel functions, integrals of, 440

Bessel inequality, 92–93

boundary conditions. See also initialvalue–boundary value problems

of the first kind. See entries atDirichlet

limitations of product method,281–282

mixed, 139

potential equation. See potentialequation

of the second kind. See entries atNeumann

of the third kind, 139wave equation, 217–220, 229

boundary value problems, 26–34. Seealso initial value–boundary valueproblems

Fourier series applications with,119–120

Green’s functions, 23, 43–49potential equation, 256–257singular, 38–41

boundedness, 40–41Boussinesq equation, 211Brownian motion, 204buckling of a column, 32–34Burger’s equation, 209

Ccable, hanging, 26–29calculus, 436–438cantilevered beam, 226car antenna vibrations, 226catenaries, 35Cauchy–Euler equation, 6–7, 38, 277Cesaro summability, 131chain rule, 232

495

496 Index

characteristic equation, 3, 10characteristics, method of, 246classifications of partial differential

equations, 280–282coefficients of Fourier series. See Fourier

seriescolumn buckling, 32–34complementary error function, 200, 202complex coefficients, potential equation

analysis, 284complex Fourier coefficients, 113–115

sampling theorem, 121–124conditions. See boundary conditions;

initial conditionconduction of heat. See heat conduction

problemsconservation of energy, law of, 135–136continuity behavior, 73–76convection, 170–174. See also heat

conduction problemsconvergence

expansions in series ofeigenfunctions, 182

Fourier series, 73–77, 124in the mean, 93–94uniform, 79–83

cooling, Newton’s law of, 30, 139cooling fins, 40–41cosine function, 66–68

Fourier cosine integralrepresentation, 109–110

hyperbolic, 4integrals of, 439

critical radius, 42cutoff frequency, 121

Dd’Alembert’s method (traveling wave),

227–231, 252damping term, 117delta functions, 113diffusion equations. See heat

conduction problemsdiffusion of sulphur dioxide, 55–56,

192–193diffusivity, 141dimensions of heat flux, 135

Dinac’s delta function, 113Dirichlet conditions, 139Dirichlet’s problem, 256. See also

potential equationin a disk, 275–279in a rectangle, 259–269soap films, 283

discontinuity, 74disk, potential equation in, 275–279

Eeigenfunctions, 158–159

expansions in series of, 181–182orthogonality of, 175–177

eigenvalue problems, 34, 158–159estimating eigenvalues for wave

equations, 236–239one-dimensional wave equation, 234singular, 189Sturm–Liouville problems, 178–179

expansions in series ofeigenfunctions, 181–182

generalizations on heat conductionproblems, 184–187

one-dimensional wave equation,234

elliptic equations, 281endpoints of periodic extensions, 76–77energy conservation, law of, 135–136enzyme electrodes, 212–214equilibrium problems. See steady-state

problemserror function, heat conduction

problems, 199–202even functions, 67

continuity behavior of, 83extensions of functions, 69–71

exponential functions, integrals of,438–439

exponential growth, 2extensions of periodic functions, 65–71

endpoints of, 76–77uniform convergence, 82–83

Ffast Fourier transform (FFT), 124Fick’s law, 55, 141

Index 497

finite Fourier series, 91–94first-order equations

homogeneous, 1–2nonhomogeneous, 20–21

Fisher’s equation, 252Fitzhugh–Nagumo equations, 239–244fixed end temperatures (heat equation),

149–155flat enzyme electrodes, 212–214flow (fluid), 284–285, 289fluid flows, 284–285, 289Fokker–Planck equation, 205forced vibrations of strings, 232forced vibrations system, 17–20forcing function, 117, 226Fourier integrals, 106–111, 124, 190,

194applications of, 117–123coefficient functions, 108complex coefficients. See complex

Fourier coefficientsFourier transforms, 115Fourier’s single integral, 112–113history of, 124representational theorem, 108wave equation in unbounded regions,

239–244Fourier series, 62–63, 124

applications of, 117–123arbitrary periods, 64–65complex coefficients. See complex

Fourier coefficientsconvergence, 73–77

proof of, 95–99uniform convergence, 79–83

cosine integral representation, 109history of, 124means of, 90–94numerical determination of

coefficients, 100–104operations on, 85–89periodic extensions, 65–71


potential in rectangle, 260–261sine integral representation, 109

Fourier transforms, 115

Fourier’s law, 30Fourier’s method (separation of

variables), 150, 166–167freezing lake, temperature of, 204frequencies of vibration, 223–224, 234functions. See specific function by name

GGaussian probability density function,

203general solutions

boundary value problems, 26homogeneous differential equations,

158nonhomogeneous linear equations,

15one-dimensional wave equation, 228second-order homogeneous

equations, 3second-order linear partial

differential equations, 205,280–281

generalized rectangles, 281generation rate functions, 141Gibbs’ phenomenon, 82Green’s functions, 23, 43–49groundwater flow, 52, 211–212

Hhalf-range extensions, 70–71hanging cable system, 26–29harmonic functions, 255. See also

potential equationheat conduction problems, 29–31,

135–206, 280convection, 170–174cooling fins, 40–41derivation of, 135–141different end conditions (example),

157–161error function, 199–202fixed end temperatures (example),

149–155generalizations on, 184–187insulated ends (example), 157–161radial heat flow, 39–40steady-state temperatures, 143–147

498 Index

higher-order equations, homogeneous,9–11

homogeneous boundary conditions,as requirement, 282

homogeneous linear equations, 1–11first-order, 1–2higher-order, 9–11second-order, 2–9

hyperbolic equations, 281hyperbolic functions, 4, 436, 438–439

Iinfinite intervals, 40–41infinite rods, 193–197initial conditions. See also initial

value–boundary value problemswave equation, 217, 220–221, 228

initial value–boundary value problems,140

heat conduction problems, 138. Seealso heat conduction problemsconvection, 170–174different end conditions

(example), 163–166fixed end temperatures (example),

149–155generalizations on, 184–187infinite rods, 193–197insulated ends (example), 157–161semi-infinite rods, 184–187

wave equation, 215–247, 280d’Alembert’s method, 227–231,

252estimating eigenvalues for, 236–239frequencies of vibration, 223–224,

234one-dimensional, in general,

233–235in unbounded regions, 239–244vibrating string problem, 215–224

insulated bars, 157–161. See also heatconduction problems

insulated surfaces, 139integrals, table of, 438–440integro-differential boundary value

problems, 54irrotational vortex, 291

Jjump discontinuities, 74

Kkryptonite, 42

LLake Ontario, 105Lake Placid, 105Laplace’s equation. See potential

equationLaplacian operator, 256–257law of conservation of energy, 135–136law of cooling (Newton), 30, 139law of radiation (Stefan–Boltzmann),

142left-hand limits, 73–74Legendre polynomials, integrals of, 440level curves, 261–262L’Hospital’s rule, 98linear density, 216linear differential equations

homogeneous, 1–11first-order, 1–2higher-order, 9–11second-order, 2–9

nonhomogeneous, 14–23Fourier series applications with,

117–119undetermined coefficients, 16–20variation of parameters, 20–23

linear operations, 140linear partial differential equations

general form, 205heat. See heat conduction problemspotential. See potential equationwave. See wave equation

linearly independent solution, 3

MMassena, New York, 132mass–spring–damper system, 5

forced vibrations system, 17–20maximum principle, 255, 278mean error, 90–94

Index 499

mean value property, 278periodic functions, 61

membrane displacement, 255. See alsopotential equation

method of characteristics, 246mixed boundary conditions, 139

NNeumann conditions, 139Neumann’s problem, 256, 290. See also

potential equationNewton’s law of cooling, 30, 139nonhomogeneous linear equations,

14–23Fourier series applications with,


nonremovable discontinuities, 74–75,79

normal probability density function,203

normalized eigenfunctions, 183normalizing constants, 183nuclear fuel rods. See heat conduction

problemsnumerical determination of Fourier

coefficients, 100–104

Oodd functions, 67

continuity behavior of, 83extensions of functions, 69–71

ODEs (ordinary differential equations),1–51

boundary value problems, 26–34Green’s functions, 43–49homogeneous, 1–11

first-order, 1–2higher-order, 9–11second-order, 2–9



singular boundary value problems,38–41

one-dimensional wave equation,233–235

ordinary differential equations, 1–51boundary value problems, 26–34Green’s functions, 43–49homogeneous, 1–11

first-order, 1–2higher-order, 9–11second-order, 2–9




ordinary limits, 73–74organ pipes, 225orthogonality, 60–61, 73

of eigenfunctions, 175–177oxygen removal from aluminum nitride,

156

Pparabolic equations, 281Parseval’s equality, 92–93partial differential equation

classifications, 280–282particular solutions, nonhomogeneous

linear equations, 15–23penetration of heat into earth, 192periodic functions, 59–63

arbitrary periods, 64–65extensions of, 65–71


piecewise continuous functions, 75–76piecewise smooth functions, 76plate, flow past, 289Poiseuille flow, 36Poisson equation, 268–269polar coordinates, potential equation in,

256–257, 275–279polynomial solution for potential

equation, 256

500 Index

potential equation, 255–285in disk, 275–279limitations of product method,

280–282Poisson equation, 268–269polynomial solution for, 256in rectangle, 259–269soap films, 283solutions to (harmonic functions),

255in unbounded regions, 270–272

principle of superposition, 3, 10, 152wave equation and standing waves,

220probability density function, 203product method (separation of

variables), 150, 166–167limitations of, 280–282potential in rectangle, 259–261, 266

Rradial heat flow, 39–40radiation, 142radical functions, integrals of, 438rational functions, integrals of, 438Rayleigh method, 239Rayleigh quotient, 239rectangle, potential equation in,

259–269reduction or order, 8–9regular singular points, 7, 38–40regular Sturm–Liouville problems,

178–179convergence theorem, 182one-dimensional wave equation, 234

removable discontinuities, 74, 79restoring term, forcing functions, 117Revision Rule, 17right-hand limits, 73–74Robin conditions, 139rod vibrations, 253rods of heat-conducting material. See

heat conduction problems

Ssampling theorem, 121–124sawtooth function, 81–82

second-order equationsgeneral form, 205heat. See heat conduction problemshomogeneous, 2–9nonhomogeneous, 21–23potential. See potential equationwave. See wave equation

sectionally continuous functions, 75–76sectionally smooth functions, 76semi-infinite intervals, 40–41semi-infinite rods, 188–191separation of variables (product

method), 150, 166–167limitations of, 280–282potential in rectangle, 259–261, 266

sine function, 66–68Fourier sine integral representation,

109–110hyperbolic, 4integrals of, 439


singular eigenvalue problems, 189singular points, 7, 38–40soap films, 283soliton (solitary) waves, 249solutions, general

boundary value problems, 26homogeneous differential equations,

158nonhomogeneous linear equations,

15one-dimensional wave equation, 228second-order equations, 3, 205,

280–281solutions, particular, 15–23square-wave function, 75–77, 79–80standing waves, 220steady-state problems. See also potential

equationtemperature (heat conduction),

143–147convection, 170–174different end conditions

(example), 157–161fixed end temperatures (example),

149–155

Index 501

generalizations on, 184–187insulated ends (example), 157–161semi-infinite rods, 184–187,

193–197wave equation, 218, 232

Stefan–Boltzmann law of radiation, 142Stokes derivative, 250stream function, 284stresses due to thermal effects, 214string, vibrating, 215–224. See also wave

equationfrequencies of vibration, 223–224,

234one-dimensional wave equation,

233–235Sturm–Liouville problems, 178–179

expansions in series ofeigenfunctions, 181–182

generalizations on heat conductionproblems, 184–187

one-dimensional wave equation, 234sulphur dioxide, diffusion of, 55–56,

192–193superposition, principle of, 3, 10, 152

wave equation and standing waves,220

surfaces, insulated, 157–161. See alsoheat conduction problems

suspension bridge (hanging cablesystem), 26–29

symmetry of sine and cosine functions,66–68

Ttable of integrals, 438–440Taylor series, 114temperature (heat conduction)

steady-state, 143–147three-dimensional steady-state

solution. See potential equationtwo-dimensional steady-state

equation, 255term, forcing functions, 117thermal conductivity, 137thermal diffusivity, 137thermal stresses, 214transient temperature distribution, 146

fixed end temperatures, 149–155transverse displacement. See wave

equationtrapezoidal function, 125traveling wave solution (d’Alembert’s

method), 227–231, 252triangle function, 123trigonometric functions, 435trigonometric series, history of, 124trivial solutions, 150truncated Fourier series, 92

Uunbounded conditions

potential equation, 270–272wave equation, 239–244

undetermined coefficients,nonhomogeneous linearequations, 16–20

uniform convergence, 79–83

Vvariation of parameters, 20–23velocity potential function, 258, 284vibrating string problem, 215–224. See

also wave equationfrequencies of vibration, 223–224,

234one-dimensional wave equation,

233–235

Wwater hammers, 250wave equation, 215–247, 280

d’Alembert’s method, 227–231, 252estimating eigenvalues for, 236–239frequencies of vibration, 223–224,

234one-dimensional, in general, 233–235in unbounded regions, 239–244vibrating string problem, 215–224,

234whirling speeds, 55windows, 111Wronskian, 3

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